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.

Question

a) How high is the cliff? i got -29.2 m but my answer is wrong
b) What was the maximum height of the ball? im not sure how to solve this part

please help, thanks in advance

Answer 1

Vi = 28 sin 60 =24.25 m/s

v = Vi - 9.8 t
h = Vi t - 4.9 t^2
= 24.25(3.3) - 4.9(3.3)^2
= 26.7 meters high
not 29.2

when does v = 0 (top of arc)?
0 = 24.25 - 9.8 t
t = 2.47 seconds
so
h at top = 24.25(2.47) - 4.9 (2.47)^2

Answer 2

Thanks Damon,
I realized that I forgot to multiply my velocity by time (3.3).
I tried to do the next part but my answer is not right. can you double check this?

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
where am I wrong?
Thanks