a tiger leaps horizontally from a 16m high rock with a speed of 7m/s.how far from the base of the rock will he land
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If the tiger is 9.0m up then from y = 1/2*g*t^2 we can determine the time of flight ===
t = sqrt(2*y/g) = sqrt ( 2*9/9.8) = 1.36s
So she will land x = vx*t = 9.0m/s*1.36s = 12.2m
t = sqrt(2*y/g) = sqrt ( 2*9/9.8) = 1.36s
So she will land x = vx*t = 9.0m/s*1.36s = 12.2m
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Explanation:
This is required answer and the answer is 12.6 meters
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