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A ball is thrown from the ground, making an angle of 53.13 degrees with respect

ID: 2126534 • Letter: A

Question

A ball is thrown from the ground, making an angle of 53.13 degrees with respect to the vertical. In 10.2 seconds, it reaches its maximum height.

The magnitude of the initial velocity of the ball is:


a) 20 m/s

b) 60 m/s

c) 125 m/s

d) 166.6 m/s

e) none of the above


I know the answer to this one is 166.6 m/s, but I keep coming up with only 99.96 m/s and I can't figure out where to work the problem next to get the correct answer.





The total distance traveled by the ball, as measured horizontally, is:


a) 100 m

b) 510.2 m

c) 1020.4 m

d) 272.0 m

e) none of the above


I know that to get this answer I have to work in the x plane, but I need the answer to the first question in order to solve this question.





The maximum height reached by the ball is:


a) 100 m

b) 510.2 m

c) 1020.4 m

d) 272.0 m

e) none of the above


I know this one is in the y plane, and I know I'm going to have to use the angle to get the answer, I'm just not sure how to go about solving the problem.


Help with the answers worked out would be amazing, and you'd be a life saver. I have to know how to do this for a test, so any help is appreciated!

A ball is thrown from the ground, making ar degrees with respect to the vertical. In 1 reaches its maximum height.

Explanation / Answer

1. As the angle is with respect to vertical, vertical component of speed = ucos(theta) = ucos(53.13) = 0.6u

It stops going upwards after 10.2 seconds. Hence we have 0 = ucos(theta) - gt

Hence u = gt/cos(theta) = 9.8*10.2/0.6 = 166.6 m/sec


2. Horizontal speed = usin(theta) = 166.6*0.8 m/sec

Time of flight = 2*10.2 = 20.4 sec

Hence horizontal distance = 20.4*166.6*0.8 = 2718.92 m. Hence correct answer is none of the above


3. Max height is where the bass stops which is given by S = ut- 0.5gt^2

= 166.6*0.6*10.2 - 0.5*9.8*10.2^2 = 509.786 which is equal to option b.

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