Question

You stand 17.5 m from a wall holding a softball. You throw the softball at the wall at an angle of 30.5 ∘ from the ground with an initial speed of 20.5 m / s. At what height above its initial position does the softball hit the wall? Ignore any effects of air resistance.

Answer 1

Answer:

5.39 m

Explanation:

From kinematics

$s=u_x t$ where t is time, s is distance and $u_x$ is initial speed in x direction

$u_x= 20.5cos 30.5=17.6634 m/s$

$u_y= 20.5 sin 30.5=10.40454 m/s$

Now 17.5=17.6634 t

$t=\frac {17.5}{17.6634}$ =0.99 s

Using the equation of kinematics

$h=u_y t- 0.5gt^{2}$

$h=10.40454\times 0.99^{2}-0.5\times 9.81\times 0.99^{2}$

h=5.39 m