A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 36.9° above the horizontal. You can ignore air resistance. (a) At what two times is the base- ball at a height of 10.0 m above the point at which it left the bat? (b) Calculate the horizontal and vertical components of the baseball’s velocity at each of the two times you found in part (a). (c) What are the magnitude and direction of the base- ball’s velocity when it returns to the level at which it left the bat
(a) Well, the baseball must be feeling pretty high to be at a height of 10.0 m. Let's calculate when this happens. Since we're ignoring air resistance, we can use some good old projectile motion. We know the initial speed is 30.0 m/s and the launch angle is 36.9°.
To find when the baseball is at a height of 10.0 m, we can use the equation h = h0 + v0y*t - 0.5*g*t^2, where h0 is the initial height (0 m), v0y is the initial vertical velocity component, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Setting h = 10.0 m and solving for t, we get 10.0 = 0 + v0y*t - 0.5*(-9.8)*t^2. Putting this quadratic equation in standard form and solving, we find two solutions for t.
(b) To find the horizontal and vertical components of the baseball's velocity at each of the two times we found in part (a), we'll use the equations vx = v0x and vy = v0y + gt, where vx is the horizontal velocity component and vy is the vertical velocity component.
We already know v0x = 30.0 m/s (since the horizontal component of the initial velocity remains constant). We can calculate v0y using v0y = v0*sin(theta), where theta is the launch angle. Then we plug in the values of v0y and t we found in part (a) and calculate vx and vy for each time.
(c) When the baseball returns to the level at which it left the bat, its vertical velocity component vy is the same as the initial value but with opposite sign (since it's moving downwards). The magnitude of the horizontal velocity component vx remains the same because air resistance never kicked in. Therefore, the magnitude of the baseball's velocity at this point is the square root of (vx^2 + vy^2), and the direction is the angle with respect to the horizontal.
I hope my answers weren't too high-flying for you!
To solve this problem, we can use the equations of projectile motion.
(a) To find the time when the baseball is at a height of 10.0 m, we can use the following equation:
y = y0 + v0y * t - (1/2) * g * t^2
where
- y is the vertical displacement (10.0 m in this case)
- y0 is the initial vertical position (0 m since the ball left the bat)
- v0y is the initial vertical velocity (v0 * sin(θ))
- g is the acceleration due to gravity (9.8 m/s^2)
- t is the time
Simplifying the equation, we get:
10.0 m = v0 * sin(θ) * t - (1/2) * g * t^2
Using the given values:
v0 = 30.0 m/s
θ = 36.9°
We can solve this quadratic equation to find the two possible values of t.
(b) Once we have the values of t, we can calculate the horizontal and vertical components of the baseball's velocity at each time.
The horizontal component (Vx) remains constant throughout the motion and is given by:
Vx = v0x = v0 * cos(θ)
The vertical component (Vy) changes due to the acceleration of gravity. At the given times, we can use the equation:
Vy = v0y - g * t
(c) Finally, to find the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat, we can use the following equations:
v = sqrt(Vx^2 + Vy^2)
θv = arctan(Vy / Vx)
Let's solve this step by step:
(a) Solve for the two possible times:
10.0 m = (30.0 m/s) * sin(36.9°) * t - (1/2) * (9.8 m/s^2) * t^2
Rearrange the equation:
(1/2) * (9.8 m/s^2) * t^2 - (30.0 m/s) * sin(36.9°) * t + 10.0 m = 0
Now we have a quadratic equation in the form of: at^2 + bt + c = 0
where a = (1/2) * (9.8 m/s^2), b = -(30.0 m/s) * sin(36.9°), c = 10.0 m.
Using the quadratic formula, t = (-b ± sqrt(b^2 - 4ac))/(2a)
Plugging in the values, we can solve for t:
t = [ -( -(30.0 m/s) * sin(36.9°)) ± sqrt(( -( -(30.0 m/s) * sin(36.9°))^2 - 4 * (1/2) * (9.8 m/s^2) * 10.0 m) ) ] / [2 * (1/2) * (9.8 m/s^2)]
Simplifying this equation will give us the two possible values of t.
(b) Calculate the horizontal and vertical components of velocity:
For each value of t, we can calculate Vx and Vy using the following equations:
Vx = v0x = v0 * cos(θ)
Vy = v0y - g * t
(c) Calculate the magnitude and direction of the velocity:
Using the values of Vx and Vy at the time when the ball returns to the level at which it left the bat, we can calculate the magnitude (v) and direction (θv) of the velocity using the equations mentioned earlier:
v = sqrt(Vx^2 + Vy^2)
θv = arctan(Vy / Vx)
Follow these steps to solve the problem step-by-step.
To solve this problem, we can use the equations of motion to analyze the baseball's trajectory. Let's break it down step by step:
(a) To find the times when the baseball is at a height of 10.0 m above the point it left the bat, we need to consider the vertical motion of the baseball. The equation that describes this motion is:
y = y0 + v0y * t - 0.5 * g * t^2
Where:
- y is the vertical position of the baseball
- y0 is the initial vertical position (0 m, since it left the bat at ground level)
- v0y is the vertical component of the initial velocity (v0 * sin(theta))
- t is the time elapsed
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
Setting y = 10 m and solving for t, we get a quadratic equation:
10 = 0 + (30 * sin(36.9°)) * t - 0.5 * 9.8 * t^2
Simplifying this equation, we obtain:
4.9*t^2 - (15 * sin(36.9°)) * t + 10 = 0
Now we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Substituting the values from our equation into the quadratic formula, we can solve for t. This will give us two values, representing the two times when the baseball is at a height of 10.0 m above the point it left the bat.
(b) To calculate the horizontal and vertical components of the baseball's velocity at each time in part (a), we can use the equations:
vx = v0x
vy = v0y - g * t
Where:
- vx is the horizontal component of velocity
- vy is the vertical component of velocity
- v0x is the initial horizontal velocity (v0 * cos(theta))
- v0y is the initial vertical velocity (v0 * sin(theta))
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time elapsed
Substituting the initial conditions and the values of t found in part (a), we can calculate the horizontal and vertical components of the velocity at each time.
(c) When the baseball returns to the level at which it left the bat, its vertical position will be 0 m. To find the time at which this happens, we can solve the quadratic equation for y = 0, considering the vertical motion as explained in part (a). This will give us one more time value.
To find the magnitude and direction of the baseball's velocity when it returns to the level at which it left the bat, we can calculate the resultant velocity using the equation:
v = √(vx^2 + vy^2)
The direction of the velocity can be found using the arctan function:
θ = arctan(vy / vx)
Substituting the values of vx and vy at the time when the baseball returns to the level at which it left the bat, we can calculate the magnitude and direction of the velocity.
By following these steps and performing the necessary calculations, you will be able to find the answers to all the parts of the question.
Vo = 30m/s[36.9o]
Yo = 30*sin36.9 = 18 m/s.
a. Y^2 = Yo^2 + 2g*h = 18^2 + (-19.6)10 = 128
Y = 11.3 m/s. at 10m.
Y = Yo + g*T1 = 11.3
18 + (-9.8)T1 = 11.3
T1 = 0.684 s.
Y = Yo + g*T2 = 11.3
0 + 9.8T2 = 11.3
T2 = 1.15 s.
b. X = Xo = 30*cos36.9 = 24 m/s.
Y = Yo + g*T1 = 18 + (-9.8)0.684 = 11.3 m /s.
Y = 0 + 9.8*1.15 = 11.3 m/s.
c. 30m/s[36.9o].
vertical problem
Vi = 30 sin 36.9 = 18 m/s upward at t = 0
v = Vi - 9.81 t
h = Vi t - 4.9 t^2
so for part a
10 = 18 t - 4.9 t^2
solve quadratic for t one and t two
when you have t
use
v = 18 - 9.81 t
for vertical component of velocity
u = horizontal velocity = same forever = 30 cos 36.9
part c is a trick question
u is the same forever
by conservation of energy or just looking at the parabola, v down at finish = v up at start :)