What was the initial speed of the baseball?

Question

A baseball player hits a ball and gives it an initial velocity that is directed at an angle of 15 degrees above the horizontal. If the ball is then caught (at the same height) by a player that is 60 meters away, what was the initial speed of the baseball?

Answer 1

Hello Lisa.

If the ball is caught at the same height, then the projectile (the ball) has gone through it's complete range horizontally. For a projectile, horizontal speed is constant (the air resistance is neglected). So the amount of time that the ball travelled horizontally can be found as Δt = v·cos15/60. During the same amount of time, vertical velocity, which was initially v·sin15 went through the course of turning into negative, with the same magnitude. That happened under the influence of gravity, which means that the ball had a vertical acceleration of -9.8 m/s². The formula for acceleration is a = (v_f - v_i)/Δt, so you can find time as - 9.8 / (- v·sin15 - v·sin15) = 9.8/2v·si15 = 4.9/v·sin15.

 

Since that amount of time is one and the same, you can set the two equations equal to each other, which will give you:

 

v·cos15/60 = 4.9/v·sin15.

 

Cross multiply and you'll get 

 

v²·sin15·cos15 = 294

v=sqr(1176) = 34.3 m/s