Arock is thrown off a cliff at an angle of 53° with respect to the horizontal. the cliff is 100 m high. the initial speed of the rock is 30 m/s. (a) how high above the edge of the cliff does the rock rise? (b) how far has it moved horizontally when it is at maximum altitude? (c) how long after the release does it hit the ground? (d) what is the range of the rock? (e) what are the horizontal and vertical positions of the rock relative to the edge of the cliff at t = 2.0 s, t = 4.0 s, and t = 6.0 s?

(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

where

is the initial velocity of the rock

is the angle

t is the time

Requiring , we find the time at which the heigth is maximum:

The heigth of the rock at time t is given by

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

where

is the horizontal component of the velocity, which remains constant during the entire motion

t is the time

If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

Substituting

and t = 7.58 s we find:

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

And substituting the different times, we find:

1/0.14nm = 10^10/1.4 a lot

multiply 2kg by latent heat of vap