In the middle of the night you are standing a horizontal distance of 14.0 m from

Question

In the middle of the night you are standing a horizontal distance of 14.0 m from the high fence that surrounds the estate of your rich uncle. The top of the fence is 5.00 m above the ground. You have taped an important message to a rock that you want to throw over the fence. The ground is level, and the width of the fence is small enough to be ignored. You throw the rock from a height of 1.60 m above the ground and at an angle of 54.0 degree above the horizontal. What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence? Express your answer with the appropriate units.

Explanation / Answer

suppose speed is u.

then initial horizontal speed = ucos54

at the point it travels 14m in horizontal, its height must be minimum of (5-1.6) = 3.4m.

so we will calculate time taken to reach 14m and then height.

in horizontal, ucos54 * t = 14

ut = 14/cos54 = 23.82 ......(i)

in vertical,

height= (usin54)t - (9.81t^2 /2 )

3.4 = ut sin54 - 4.905t^2

4.905t^2 = (23.82)sin54 - 3.4

t = 1.80 sec

so speed u = 23.82 / t = 13.24 m/s

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