Consider a rock that is thrown off a bridge of height 16 m at an angle ? = 22° w

Question

Consider a rock that is thrown off a bridge of height 16 m at an angle ? = 22° with respect to the horizontal as shown in the figure below. If the initial speed the rock is thrown is 14 m/s, find the following quantities.
(a) The time it takes the rock to reach its maximum height.
0.535 s
(b) The maximum height reached by the rock.
17.4 m
(c) The time at which the rock lands.
2.42 s
(d) The place where the rock lands.
31.4 m
(e) The velocity of the rock (magnitude and direction) just before it lands.
magnitude

direction ° (deg)

I have gotten all of the answers but the last two (part e)
I can not figure out how to find the magnitude and direction of the rock just before it lands. I have filled 3 sheets of paper trying to solve this issue. Please help me out. So far I have tried calculating Vyf but am not sure if I should use the original y-direction velocity (=5.24 m/s) or another one, like when the rock is at its peak (Vy=0)...Please clarify and help me understand what concept I'm missing!

Explanation / Answer

e)

The horizontal velocity Vx = 14 cos22 = 12.98 m/s

The vertical velocity Vy = [(14 sin22)2 + (2)(9.8)(16)] = 18.50 m/s

The magnitude of velocity v = 22.57 m/s

direction tan = Vy / Vx = 18.50 / 12.98 = 1.425

                   = 54.94 deg

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