Stone throw, speed and angle?

Question

A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 17.0 m/s. The cliff is h = 61.0 m above a flat horizontal beach.



How long after being released does the stone strike the beach below the cliff?
_____s
With what speed and angle of impact does the stone land?
_____m/s
_____° below the horizontal

Answer 1

the force vertically acting on the stone is the gravity

The fall occurs on 61m and the time is given by

d= (1/2 )*g*t^2 so t = (2d/g)^0.5 =3.5s

the vertical component of speed is V(v) = gt= 3.5*9.81=12.3 m/s

For the angle horizontal component does not change =17 m/s

vertical is 12 so teh angle tg(angle) = 12.3/17= tg 36°

Answer 2

make certain the stone's flight into horizontal and vertical elements you at the moment have the initial speed of the stone going upwards Calculate the top it reaches till its speed is 0 (you at the moment have the optimal top of the stone) Then basically practice your regular formula to paintings some thing else out