Consider a rock that is thrown off a bridge of height 56 m at an angle ? = 22° w

Question

Consider a rock that is thrown off a bridge of height 56 m at an angle ? = 22° with respect to the horizontal as shown in the figure above. If the initial speed the rock is thrown is 12 m/s, find the following quantities:
(a) The time it takes the rock to reach its maximum height.
0.46s

(b) The maximum height reached by the rock.
57.03m

(c) The time at which the rock lands.
_______s   I know y= y0 + V0y - 1/2gt^2 but can not figure out how to calculate for t

(d) The place where the rock lands.
_______ m

(e) The velocity of the rock (magnitude and direction) just before it lands.
Magnitude Direction
______ m/s _________°

I have found a and b part but am having the worse time calculating the rest of the problem. Any help is appreciated Thanks

Explanation / Answer

Calculate c in 2 parts. i> time of flight above the rock ii> time of flight below it part i is simple for part ii ,u know the vertical downward velocity and height . Calculate time using formula for distance. total time is sum of both d>u know horizontal velocity from symmetric behaviour of projectile and u know the time ,calculate distance. e>here us Vy = Uy + Ay*t reply if any doubt

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