A boy kicks a rock off a cliff with a speed of 19.4 m/s at an angle of 52.5° above the horizontal. The rock hits the ground 5.41 s after it was kicked.
a) How high is the cliff? (60.3 m)
b) What is the speed of the rock right before it hits the ground?
c) What is the maximum height of the rock measured from the top of the cliff?
d = Vit + 1/2 at^2
vf^2 = vi^2 + 2ad
The Attempt at a Solution
a) I found this by using the equation:
d= (-19.4sin52.5)(5.41) + (0.5)(9.8)(5.41)^2
= 60.3 m, this is correct
b) I am having trouble figuring out the speed as it hits the ground. I was trying:
Vf^2 = 0 + 2(9.8)(60.3)
= 34.4 m/s, however this seems to be wrong..... what am i doing wrong?
c) I do not know how to start this part.
Somebody please help me!