Question

7) A baseball player releases a ball with an initial velocity of 35 m/s and it takes 2 seconds to reach max height. How high did the ball get? How long would it be before the other baseball player catches the ball? What is the release angle with respect to the vertical plane? How far into the field would the ball get?

Answer 1

7. Ball will take half of time of flight to reach at maximum height . 2= 7/2 =) 4= T =) T-4 Manz =) .2 Vo sino , tzo tausee ug =) Sino = 4 avo 2x9.8 22 - 056 1906 = B = Sint(0.56) = 34.050 Ans Calculation of maximum height H- Vo sinto - (35²xsin 234.05° - 35²x10.5602 2x9.8 - 19.6m. Aus Time of flight between two player = 25 + 25 = u see. Ang Release Angle o=340 with horizontal or Release Release angle angle with horizontal = 90° -0 = 90-34° = 56° Ang Horizontal distance covered by ball x= Vocoso xt = 35 cos 340 x 4 Hence we have he Answers Answere 116 m. ♡ H = 19.6m ② T=4sec ③ 2 =560 W n=116m.