A rock is thrown off a cliff at an angle of 48° above the horizontal. The cliff

Question

A rock is thrown off a cliff at an angle of 48° above the horizontal. The cliff is 135 m high. The initial speed of the rock is 24 m/s. (Assume the height of the thrower is negligible.)

(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

Initial Velocity of Rock is

Vi = 24 m/sec about 48 deg above the horizontal

Vix = 24*cos 48 deg = 16.06 m/sec

Viy = 24*sin 48 deg = 17.84 m/sec

Height of cliff = 135 m

acceleration is

ax = 0

ay = -g = -9.81 m/sec^2

Now we know that since there is no horizontal acceleration, So horizontal velocity will remain constant

So distance traveled in x-direction will be given by:

X(t) = Vix*t

At t = 2 sec

X(2) = (16.06 m/sec)*2 sec = 32.12 m

At t = 4 sec

X(4) = (16.06 m/sec)*4 sec = 64.24 m

At t = 6 sec

X(6) = (16.06 m/sec)*6 sec = 96.36 m

Now Vertical position will be given by:

y(t) = Viy*t + 0.5*ay*t^2

At t = 2 sec

y(2) = 17.84*2 - 0.5*9.81*2^2

y(2) = 16.06 m

At t = 4 sec

y(4) = 17.84*4 - 0.5*9.81*4^2

y(4) = -7.12 m

At t = 6 sec

y(6) = 17.84*6 - 0.5*9.81*6^2

y(6) = -69.54 m

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