A rock is thrown off a bridge of height 75m. at an angle theta=25 degrees with r

Question

A rock is thrown off a bridge of height 75m. at an angle theta=25 degrees with respect to the horizontal. The initial speed of the rock is 15 m/s. Find the following quantities:
a) the maximum height reached by the rock?
b) the time it takes the rock to reach its maximum height?
c) the place where the rock lands?
d) the time at which the rock lands
e) the velocity of the rock (magnitude and direction) just before it lands?

Explanation / Answer

Vx = 15 cos 25 = 13.595 m/s Vy = 15 sin25 =6.34 m/s a). v^2 -u^2 = 2as => 0 - 6.34^2 = 2*(-9.8)*s => s =2.051 m maxm height = 75+2.051 = 77.051 m b). v = u +at =>0 = 6.34 -9.8 t => t = 0.647 s c). s= ut+0.5at^2 => 77.051 = 0+ 0.5*9.8*t^2 => t = 3.965 s total time taken to reach ground = 3.965 +0.647 = 4.612 s horizontal displacement = 4.612* 13.595 = 62.7 m d). t = 4.612 s e). V'y = 0 +9.8* 3.965 => V'y = 38.857 m/s V'x = 13.595 m/s magnitude = sqrt[(13.595)^2 + (38.857)^2] = 41.167 m/s tan(theta) = 38.857/13.595 = 70.72 degrees with the horizontal

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