The horizontal velocity component remains
Vx = 18.0 m/s
during the fall to the beach.
The vertical component of velocity (measured positive upwards) is
Vy = - gt
The time t that it takes to reach the beach when falling from height H is given by solving
H = (g/2) t^2
Therefore t = sqrt (2H/g) = 3.19 s
That answers your second question. Use that value of t to compute Vy at impact and from that and Vx get the angle of impact.
The speed of impact is sqrt[Vx^2 + Vy^2]
Multiple part question…..
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0m above a flat, horizontal beach. With what angle of impact does the stone land (in degrees).
A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18 m/s. the cliff is 50 m above a flat, horizontally beach. how long after being released does the stone strike the beach below the cliff?
A student atands at the edge of a cliff and throws a stone horizontally over the edge with a speed of 18.0 m/s. The cliff is 50.0 m above flat, horizontal beach. With what speed of impact does the stone land?
1 answer
1 answer