Question

A man stands on the roof of a 10.0 m -tall building and throws a rock with a velocity of magnitude 24.0 m/s at an angle of 34.0 degrees above the horizontal. You can ignore air resistance.

Calculate the magnitude of the velocity of the rock just before it strikes the ground.

&

Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground

Answer 1

\[\alpha=34\]
V=24m/s
h=10m
g=9.81
time from the building to hmax:


\[t=( V \times \sin \alpha)/g\]

now hmax=
\[V \times \sin \alpha \times t - (g \times t ^{2})/2=hmax\]


\[(V^{2} \times \sin^{2}\alpha)/(2\times g)=hmax\]
so the horizontal distance:
\[x=V \times \cos \alpha \times(V \times \sin \alpha )/g + V \times \sqrt{(2 \times (h+hmax)/g)}\]
\[x \approx48,59 m\]

velocity before strike:

\[v =\sqrt{g \times \sqrt{(2 \times (h+hmax)/g)} + V^{2} \times \cos^{2} \alpha}\]
\[v \approx20,37m/s\]
I could make a mistake in calculations
sorry for quality i'm newbe