Question

A ball player hits a home run, and the baseball just clears a wall 21.0 m high located 130.0 m from home plate. The ball is hit at an angle of 35.0° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.0 m above the ground.

Answer 1

Answer:

$v = 41.7 m/s$

Explanation:

As we know that the position of the ball is at initially height 1 m from the ground

so now we will have

$y = y_o + v_y t + \frac{1}{2}at^2$

now we know that

$a = -9.81 m/s^2$

$y_o = 1 m$

also we have vertical component of the velocity is given as

$v_y = vsin35$

now we have

$21 = 1 + (vsin35) t - \frac{1}{2}gt^2$

also in x direction we have

$x = v_x t$

$130 = (vcos35) t$

so we have

$21 = 1 + (vsin35)(\frac{130}{vcos35}) - 4.905 t^2$

$4.905 t^2 - 91.03 - 1 + 21 = 0$

$t = 3.8 s$

now we have

$v = \frac{130}{t cos35}$

$v = \frac{130}{3.8 cos35}$

$v = 41.7 m/s$