# ) A batter hits a baseball so that it leaves the bat at speed

Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2. (a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the Iwrizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.

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1. Question:

A batter hits a baseball so that it leaves the bat at speed
Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2.

(a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s.

SOLUTION FOR A:

POSITION OF THE BALL:
In order to find the position of the ball, split up the Vo component into Vox and Voy. That gives you:

Vox = 37.0cos(57.1) =22.2 m/s
Voy = 37.0sin(57.1) = 29.6 m/s

Once you solve for those, your goal is to find the final positions of the x and y coordinates, and you use the kinematic equation for position when given time, acceleration, and velocity: x= xo+vt+1/2at^2 -->

Solving for X final:
Vox = 22.2 m/s

Xf=Xo+vt+1/2at^2, and intial position of x is 0 and since there is no acceleration in the X direction get rid of that last term (1/2at^2). So now you simplify to get the equation Xf=vt, which is just Xf= Vox *t --> Xf = 22.2(2)= 44.4m

Solving for y final:
Voy = 29.6 m/s

Yf=Yo+Vyo*t+1/2at^2, now again, Yo is 0 and so you simplify this equation down to Yf=Voy*t+1/2*at^2, and the reason you don't get rid of the last term for Y is because you have gravity, and gravity is given as g=9.8=a. Yf=29.6(2)+(1/2)(-9.8)(2^2). This is equal to 39.6.

Your final answer for position is (44.4, 39.6)

SOLVING FOR MAGNITUDE OF VELOCITY FOR THE BALL AT T=2:

Magnitudes are just the distances found by s=sqrt(x^2+y^2), and the magnitude of velocity is V(magnitude)=sqrt(Vxf^2+Vyf^2). We're solving for V(mag), you need to fine Vxf and Vyf.

Use the Vf=Vo+at:

for the Vxf:
Vxf=Vxo+at, cancel out at because once again, you don't have acceleration in the X component. You therefore get, Vxf=Vxo, and Vxo=22.2m/s, so Vxf=22.2m/s.

For the Vyf:
Vyf=Vyo+at, you have acceleration of gravity g=9.80m/s^2, and t=2, and plugging Vyo=29.6m/s, Vyf=29.6+(-9.8)(2), and that equals 10. Vyf=10m/s.

Vxf=22.2m/s.
Vyf=10m/s.

Now that you have Vxf and Vyf at t=2, you can solve for the magnitude of the velocity at t = 2 by V(magnitude)=sqrt(Vxf^2+Vyf^2) --> V(mag)= sqrt(22.2^2+10^2)=24.4 m/s

Magnitude of Velocity at time = 2 seconds is 24.4m/s

SOLVING FOR DIRECTION AT TIME 2 SECONDS:

The direction (or angle) is found by tan^-1(y/x), for the velocity it's therefore tan^-1(Vyf/Vxo)

Vxf=22.2m/s.
Vyf=10m/s.

Tan^-1(10/22.2)= 24.3º is your angle

The direction of velocity is 24.3º from the horizontal at t=2 secs.

Position: (44.4, 39.6)
Magnitude of Velocity: 24.4m/s
Direction of Velocity: 24.3º

SOLVING FOR B:

(b) Find the time when the ball reaches the highest point of its flight and find its height h at this point.

To solve for the time when the ball has reached it's highest point, Vfy must equal 0, because if it's not longer moving upwards in the y direction that means it's reached it's max height. Use the second kinematics equation to find t given the initial and final velocities, and the acceleration due to gravity : Vyf=Vyo+at, substituting our knowns, a=-9.8, Vyf=0, and Vyo=29.6, we have 0=29.6+-9.8t, and therefore t=-29.6/-9.8, resulting in t=44.7 m. The time it takes for the ball to reach the highest point is 3.02 m.

t= 3.02 seconds

Solving for height, requires the first kinematics equation x=xo+vt+1/2at^2, and this turns into Yf=Yo+Vyo*t+(1/2)a*t^2, your knowns:
t=3.02
a=-98
Yo=0
Vyo=29.6

Yf=0+29.6(3.02)+(1/2)(-9.8)(3.02^2) =>
Yf=44.7 m
The height h at the highest point is 44.7 meters.

Time: 3.02 seconds
Height: 44.7 meters.

SOLVE FOR C:

(c) Find the horizontal range R-that is, the horizontal distance from the starting point to where the ball hits the ground.

Since projectile motion in this problem will be symmetrical in distance, you can solve for the x distance traveled when it reaches it's maximum height at t=3.02 seconds using the first kinematic equation Xf=Xo+Voxt+1/2at^2, that is X=22.2(3.02), that gives you Xf=67.044, now multiply this by 2 to get the whole range, 67.044*2= 134

Range: 134 meters

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2. Vo = 37 m/s
Theta = 53.1
Xo = 0 m i
Yo = 0 m j
ax = 0 m/s^2 i
ay = -9.8 m/s^2 j

Vox = Vo * costheta
= 37[cos(53.1)]
= +22.2 m/s i

Voy = Vo * sintheta
= 37[sin(53.1)]
Voy = +29.6 m/s j

x1 = +44.4 m i

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3. (a) x = 37.0 cos53.1 * t = 22.22 t
y = 37.0 sin53.1 t - (g/2) t^2
= 29.59 t - 4.90 t^2
Vx = 22.22
Vy = 29.59 - 9.8 t
angle to horizontal = arctan Vy/Vx
Plug in t = 2 s

(b) Compute y at time when Vy = 0
t = Vyo/g = 3.02 s
y = (Vyo)^2/(2g)

(c) R = 2 Vo^2 sin53.1*cos53.1/g
= 0.96 Vo^2/g

Vo is the launch velocity and Vyo is the y component of the launch velocity

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4. 24.23m/s, 23.21 degree

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