A ball is thrown so that the motion is defined by the equations x= 5t and y= 2+6t-4.9t^2, where x and y are expressed in meters and t is expressed in seconds. Determine (a) the velocity at t=1s, (b) the horizontal distance the ball travels before hitting the ground
the velocity is constant in the x-direction ... dx/dt = 5 m/s
the velocity in the y-direction is time dependent ... dy/dt = 6 - 9.8 t m/s
(a) plug in t = 1 , and add the x and y velocity vectors
(b) set y=0 and solve for t , plug the result into the x equation
a. 5.88 m/s
b. 7.5 m
(a) To determine the velocity at t=1s, we need to find the derivatives of the equations x= 5t and y= 2+6t-4.9t^2 with respect to time:
- The derivative of x=5t with respect to t is dx/dt = 5.
- The derivative of y=2+6t-4.9t^2 with respect to t is dy/dt = 6 - 9.8t.
At t=1s, the velocity is given by the derivatives' values:
- Velocity in the x-direction: dx/dt = 5 m/s.
- Velocity in the y-direction: dy/dt = 6 - 9.8(1) = 6 - 9.8 = -3.8 m/s.
So, the velocity at t=1s is 5 m/s horizontally and -3.8 m/s vertically.
(b) The ball hits the ground when y=0. Setting the y equation 2+6t-4.9t^2 equal to zero:
2+6t-4.9t^2 = 0.
We can solve this quadratic equation to find the times when the ball hits the ground. However, for the purpose of determining the horizontal distance traveled before hitting the ground, we only need to look at the x-component of motion.
The horizontal distance traveled (x-distance) can be found by substituting t=0 into the x equation since the ball is thrown from the origin:
x = 5t = 5(0) = 0.
Therefore, the horizontal distance the ball travels before hitting the ground is 0 meters.
(a) To determine the velocity at t = 1s, we need to find the derivatives dx/dt and dy/dt with respect to time.
Given:
x = 5t
y = 2 + 6t - 4.9t^2
Taking the derivative of x with respect to t:
dx/dt = d(5t)/dt = 5
Taking the derivative of y with respect to t:
dy/dt = d(2 + 6t - 4.9t^2)/dt = 6 - 9.8t
At t = 1s, substituting t = 1 into dx/dt and dy/dt:
Velocity at t = 1s:
v = dx/dt = 5
v = dy/dt = 6 - 9.8(1)
v = 6 - 9.8
v ≈ -3.8 m/s
Therefore, the velocity at t = 1s is approximately -3.8 m/s.
(b) To find the horizontal distance the ball travels before hitting the ground, we need to determine when the ball hits the ground. The ball hits the ground when y = 0.
Given:
y = 2 + 6t - 4.9t^2
Setting y = 0:
0 = 2 + 6t - 4.9t^2
4.9t^2 - 6t - 2 = 0
We can solve this quadratic equation using the quadratic formula, which states that for an equation of the form ax^2 + bx + c = 0, the solutions are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In our case, a = 4.9, b = -6, and c = -2.
Using the quadratic formula:
t = (-(-6) ± √((-6)^2 - 4(4.9)(-2))) / (2(4.9))
t = (6 ± √(36 + 39.2)) / 9.8
t = (6 ± √75.2) / 9.8
Now we can solve for t:
t ≈ 2.085 s (rounded to three decimal places)
The ball hits the ground at approximately t = 2.085 s.
To find the horizontal distance traveled, substitute this value of t into the equation for x:
x = 5t
x ≈ 5 * 2.085
x ≈ 10.425 m
Therefore, the horizontal distance the ball travels before hitting the ground is approximately 10.425 meters.
To determine the velocities and the horizontal distance traveled by the ball, we need to differentiate the equations for position with respect to time.
Given:
x = 5t (Equation 1)
y = 2 + 6t - 4.9t^2 (Equation 2)
(a) The velocity at t=1s:
Velocity is the rate of change of displacement with respect to time. To find the velocity, we need to differentiate the equations for x and y with respect to t.
Differentiating Equation 1:
dx/dt = d(5t)/dt = 5 (since the derivative of t is 1)
Differentiating Equation 2:
dy/dt = d(2 + 6t - 4.9t^2)/dt
= 6 - 9.8t (since the derivative of t^2 is 2t)
Now, substitute t=1s into the derivative to find the velocity at t=1s:
dy/dt = 6 - 9.8(1)
= 6 - 9.8
= -3.8 m/s
Therefore, the velocity at t=1s is -3.8 m/s (negative value indicates the downward direction).
(b) The horizontal distance the ball travels before hitting the ground:
The ball hits the ground when y=0 because the position along the y-axis represents the height.
Substitute y=0 into Equation 2 and solve for t:
0 = 2 + 6t - 4.9t^2
4.9t^2 - 6t - 2 = 0
This is a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 4.9, b = -6, and c = -2. Substitute the values into the formula and solve for t:
t = (6 ± √((-6)^2 - 4(4.9)(-2))) / (2 * 4.9)
t = (6 ± √(36 + 39.2)) / 9.8
t = (6 ± √75.2) / 9.8
Since time cannot be negative, we need to take the positive root:
t = (6 + √75.2) / 9.8
Now, substitute this value of t back into Equation 1 to find the horizontal distance traveled by the ball:
x = 5t
= 5[(6 + √75.2) / 9.8]
Calculating this expression will give you the horizontal distance traveled by the ball before hitting the ground.