A rock is thrown off a cliff at an angle of 46° above the horizontal. The cliff

Question

A rock is thrown off a cliff at an angle of 46° above the horizontal. The cliff is 145 m high. The initial speed of the rock is 25 m/s. (Assume the height of the thrower is negligible.)

(a) How high above the edge of the cliff does the rock rise (in m)?

(b) How far has it moved horizontally when it is at maximum altitude (in m)?

(c) How long after the release does it hit the ground (in s)?

(d) What is the range of the rock (in m)?

(e) What are the horizontal and vertical positions (in m) of the rock relative to the edge of the cliff at

t = 2.0 s,

t = 4.0 s,

and

t = 6.0 s?

(Assume the +x-direction is in the horizontal direction pointing away from the cliff, the +y-direction is up towards the sky, and x = y = 0 at the point from which the rock is thrown.)

x(2.0 s)=

y(2.0 s)=

x(4.0 s)=

y(4.0 s)=

x(6.0 s)=

y(6.0 s)=

Explanation / Answer

(A) at maximum height , vfy -= 0

vfy^2 - v0y^2 =2 a d

0^2 - (25 sin46)^2 = 2(-9.8) h

h = 16.5 m

(B) vf= v0 + a t

0 = 25 sin46 - 9.8 t

t= 1.835 sec

d_x = (23 cos46)t = 29.3 m

(c) yf - yi = v0y t + ay t^2 /2

0 - 145 = 25 sin46 t - 9.8 t^2 /2

4.9 t^2 - 25 sin46 t - 145 = 0

t = 7.58 sec

(D) range = (25 cos46)t = 131.6 m

x = 25 cos46 t

y = 25 sin46t - 4.9t^2

x2 = 34.7 m

x4 = 69.5 m

x6 = 104.2 m

y2 = 16.4 m

y4 = -6.5 m

y6 = - 68.5 m

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