A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0 m/s and at an angle of 36.5 m/s above the horizontal. You can ignore air resistance.

A. At what two times is the baseball at a height of 10.0m above the point at which it left the bat?
Enter your answer as two numbers, separated by a comma, in the order t1,t2 , where t2>t1. ans in seconds

B. Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A.
Enter your answer as two numbers, separated by a comma, in the order v1,v2 . ans in m/s.

C. Calculate the vertical component of the baseball's velocity at each of the two times you found in part A.
Enter your answer as two numbers, separated by a comma, in the order v1,v2 . ans in m/s.

Explanation / Answer

It doesnt say anything about hozizontal distance travelled so just look at the vertical component of the velocity. First get the vertical component of the velocity: v=28sin(36.9)=16.8 (you can draw a diagram of horizontal, vertical and total velocity and use trigonometry) Now newton's second law is F=ma which can be integrated for a constant (gravitational) force to get: x=1/2at^2+vo*t+x0 where vo is initial vertical velocity and xo is initial height. Note everything is being considered in the up down direction cos this is the direction of gravitational force. a=-9.81m/s^2 which is acceleration due to gravity, it is negative because it points down not up. hence x=8.5 =1/2*9.81*t^2+28sin(36.9)*t+0 can be rearrange to get 0= - 4.9t^2+16.8 t-8.5

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