Physics problem help?

Question

A rock is thrown from a 21 m. high cliff at 21 m/sec at an angle of 53 degrees. How long does it take for the ball to land? How far away from the cliff does the ball land? (the range of the ball)

*ball is thrown, not rock

Answer 1

Assume no friction
Next compute the horizontal and vertical components of initial velocity. I interpret the question to state the 53 degrees to be a downward angle from the horizontal plane.

So the vertical =Sine(53) * 21
horizontal = Cos(53)*21

Also, gravity will act on the ball at 9.8m/s^2 in the vertical direction only.

We know the height to be 21 m
Initial velocity plus the average increase due to gravity times time gives the distance:

21=(sine(53)*21 +9.8/2 * t)* t
0 = 4.9 * t^2 + sine(53) * 21 * t - 21
sole for t ( the positive one since this a parabola)

4.9 * t^2 +16.77*t - 21

t=.975 secs

The range is the horizontal component of velocity times t

range = .975 * Cos(53) * 21

range = 12.32 m

If the rock/ball is thrown at an upward angle, the solution is similar. First find the point of apogee by computing the vertical velocity as initial - 1/2 g t1 = 0
t1 = 3.42 secs


then compute the distance traveled in the vertical to find the height. apogee= 28.7 m the object will then fall that distance plus 21m in the vertical acted upon by gravity until it strikes the ground.

1/2 g t2 ^2 = 21 + apogee
t2 = 3.18 secs

Add t1 + t2 to compute the time of travel and multiply by Cos(53) * 21

range = 81.4 m

j

Answer 2

You just consider the cliff is forming a right angle triangle's hypotenuse h. The lower angle is 53 degrees so the upper one is 90 degrees-53 degrees=37 degrees.
Using the trigonometric laws, you get:
sin 53degrees=21/h
h=21/(sin53)
h=26.4
You now have the distance the ball must travel. You can compute the time it will take to land:
26.4/21=1.26 seconds

I'm better in math than in physics, but I don't think you are able to calculate the distance at which the ball will fall from the cliff.

Answer 3

Apply S = ut + 1/2at^2
{s = displacement, u=starting velocity, a= acceleration, t=time}
for vertical motion (upwards)
-21 = 21Sin Q. t +1/2(10)t^2
-21 = 21 SinQ - 5t^2
5t^2 - 21.Sin53 - 21=0
t = 4.325 seconds
.
Apply S = ut + 1/2at^2 for horizontal motion
s = 21 Cos53.t + 1/2 (0) t^2
s = 54.66 meters

Answer 4

Well, the ball doesn't exist, because you threw a rock.