You\'re standing on a cliff overlooking the ocean when you suddenly feel the nee

Question

You're standing on a cliff overlooking the ocean when you suddenly feel the need to throw a rock over the cliff into the ocean. You calculate the height from your point of view to the water to be H = 225m. You notice that between you and the ocean there is a hump h = 150m in height above the water that you must clear as you throw the rock. (a) If you throw the rock at an angle of theta = 18.5degree below the horizontal, with what initial velocity V_0 must it be thrown in order to overcome/surpass the hump h and reach the ocean? (b) What is the total distance from where you stand on the x-axis to the point the rock enters the water?

Explanation / Answer

H = 225 m
h = 150m
d = 206 m
theta = 18.5 degrees
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let u be the launching velocity.
then horizontal component of u , u_x = u cos 18.5
vertical component of u, u_y = u sin 18.5
time taken to travel distance d , t = d/u cos18.5 = 206/u*0.95   = 216.84/u
for vertical fall of stone , time of fall:
H = usin18.5 * t + 0.5gt^2
225 = 216.84 /u * u sin 18.5   + 0.5 *9.81 *(216.84/u)^2
225 = 216.84 sin18.5 + 0.5*9.81*(216.84/u)^2
225 = 68.8 + 230631.07/u^2
u = 38.425 m/s

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