Hi there! Thank you for the opportunity to help you with your question!
In this problem, we have a ball travelling under the influence of gravity, with an initial height y0 = 1.5 (in the vertical direction), and moving without any forces (in the horizontal direction). The initial velocity vector v0 = (v0x, v0y). We can calculate the components by noting that the vector forms an angle of 48 degrees, therefore, v0x = v0*cos(48)=12.71 m/s ; and v0y = v0*sin(48) = 14.12 m/s
The equations of motion are given by:
x = v0x*t
y = y0 + v0y*t - gt^2/2
We actually only need the vertical equation to solve these two questions. The time the ball is in the air is equal to the time it takes for it to climb up, and fall back down to a height of y=1.5 m. Plugging in all the values:
1.5 = 1.5 + 14.12t -9.8t^2/2
0 = t(14.12-4.9t)
This means that the ball was at y = 1.5 m when t=0 (we knew that from the beginning of the problem) but also at t = 14.12/4.9 = 2.882 seconds.
There are two ways to calculate the vertical velocity component right before Sarah catches it. One is to use the formula:
vy =v0y - 9.8*t
vy = 14.12 - 9.8*2.882
vy = -14.12 m/s
Or, we could have noticed that the problem is completely symmetric! Julie threw the ball with a vertical component of 14.12 m/s upwards, and when it arrives at the same height, to her friend Sarah, the ball is just going down, with the same speed.
Please let me know if you need any clarification. Always glad to help!
After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 14 m above the ground, and takes 2.44 s to get directly over Julie's head.
What is the speed of the ball when it leaves Sarah's hand?
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