A child standing on the edge of a sheer cliff wall throws a rock down towards the ground below. The rock is thrown 67° below the horizontal at a speed of 22 m/s & lands 43 m from the base of the cliff wall. Ignore air drag.

Question

A.) Determine how long the rock was in the air.
  s

B.) Determine how high the cliff wall is.
  m

C.) Determine how fast the rock was going upon impact.
  m/s

D.) Determine the angle of impact.
  °

Flavia · Accepted Answer

a) Use delta x= V * delta t
43 m= 22 m/s*cos 67 *delta t
delta t= 5.00 s
b) Use delta x= Vo*delta t + 1/2 * a* delta t ^2
delta x= 22 m/s*sin(-67)(5.00s)+ 1/2 *-9.81 m/s^2 * (5.00s)^s
delta x= -223.88m
Since the rock went down 223.88m, the cliff is 223.88 m high.
c)Use Vf^2-Vo^2= 2a*deltax
Vf= sqrt (2a*deltax+Vo^2)
Vf= +- sqrt |(2*-9.81m/s^2*-223.88m+(22m*sin(-67))^2)|
Vf=-69.301 m/s
However, this is only the y component. Use Pythagorean theorem to find the final velocity:
Final velocity = sqrt (-69.301m/s ^2 + 22*cos(-67) ^2)
Final velocity=69.832 m/s
d) use arctan to find the angle of impact. arctan (-69.301/(22*cos(-67))
Angle= -82.929
Hope this helps!