The next three questions pertain to a boy who is throwing tennis balls across a

Question

The next three questions pertain to a boy who is throwing tennis balls across a flat ground. All the balls are thrown from h0 = 1.5 m above the ground.
In one case, he throws the ball at an angle 1 with respect to ground and with a speed v0= 13 m/s. At time t1 = 1.3 s, the ball is h1 = 2.4 m above the ground.

What is v0y, the initial velocity in the y-direction?

How far in the x-direction has the ball travelled at t1?

In a second throw, the boy is changing the initial speed of the ball, and he throws the ball at angle, 2, that is larger than 1. But, just as in the previous case, the ball is 2.4 m above the ground at time t1 = 1.3 s. How does v0x,2, the x-component of the ball's initial velocity in the second throw, compare to v0x,1, the the x-component of the ball's initial velocity in the first throw?

v0x,2 < v0x,1

v0x,2 = v0x,1

v0x,2 > v0x,1

Explanation / Answer

Case 1 -
Initial Horizontal Velocity = vo * cos(1) = 13*cos(1)
Initial Vertival Velocity = vo * sin(1) = 13*sin(1)

Vertical distance travelled in time t1 = 2.4 - 1.5 = 0.9m

s = u*t + 0.5*g*t^2
0.9 = 13*sin(1) * 1.3 - 0.5*9.8*1.3^2
1 = 32.9 degree

Initial velocity in the y-direction = 13*sin(32.9)
Initial velocity in the y-direction = 7.06m/s

Ball travelled in x direction in time t1 = 13*cos(32.9) * 1.3m
Ball travelled in x direction in time t1 = 14.19m

Case 2 -

As 2 > 1, Therefore -
Cos(2) < cos(1)
Therefore,
Horizontal Component of velocity in 2nd case < Horizontal Component of velocity in 1st case
v0x,2 < v0x,1

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