Question

A ball is thrown horizontally from window that is 30m high with a velocity of 12m/s.find how long the ball takes to reach the ground? how far from the building it lands?and how fast the ball hits the ground?

Answer 1

use the formula here
d=vcosA(vsinA+sqrt[(vsinA)^2+2gyo]/g where A=angle at w/c the projectile is launch,
here A=0 therefore
d=12cos0(0+sqrt[0+2(9.8)30]/9.8=29.69m approx
time of flight t=d/(vcosA)=29.69/12=2.47sec
it hit the ground at a velocity v=sqrt[v^2 +(gx/vcosA)^2]
v=sqrt[(12)^2 +((9.8)(29.69)/12)^2]=27.05 m/s

Answer 2

We don't really need the cos here.

\[s = V_{0}(t) + { 1 \over 2 } g t^2\]
\[30 = 0 + { 1 \over 2 } \times 9.81 \times t^2\]
\[t^2 = {30 \over 9.81}\]
\[t \approx 1.75 s\]

\[12 ms^{-1} @ 1.75s = 20.98 meters\]
\[V _{vertical}= 1.75 \times 9.81 \approx 17.17 ms^{-1}\]