Question

A ball is thrown toward a cliff of height h with a speed of 28m/s and an angle of 60∘ above horizontal. It lands on the edge of the cliff 3.3s later.
a) How high is the cliff? i got 26.7 m
b) What was the maximum height of the ball? my answer here was 30 m
C) What is the ball's impact speed?
I found fall time 3.3-2.47=0.83s
then I used this formula V1=V0+gt = 0+9.80*0.83 = 8.13m/s

But my answer is wrong. Can youoplease help me?
Thanks

Answer 1

Cliff height is correct
Max height is correct (I got 29.7m)
Ball's impact speed isn't right.

you need to find Vx and Vy at time 3.3s

Vx = Ux - 0*t (gravity doesn't act in the horizontal direction)
Vy = Uy - g*t (gravity acts downwards so -ve g)

So we get Vx = Ux = 14m/s
and Vy = 14*sqrt(3) - 9.8*3.3 = -5.15m/s
That minus just means it's falling at that time.

To find the speed at that time
V^2 = (Vx)^2 + (Vy)^2
V^2 = 222.52
V = 14.9m/s