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

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 40.0 ∘above the horizontal.

What minimum initial speed must the rock have as it leaves your hand to clear the top of the fence?

For the initial velocity calculated in the previous part, what horizontal distance beyond the fence will the rock land on the ground?

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Answer #1

Angle of projection θ-40 Horizontal di splacementx- 14 m Vertical displacement y = 5-1.6-3.4 m Let u be the minimum velocity of projection Equation of motion of a projectile is 9.8 -34 = tan400 (14)- man (14) 2u cos 40 1636.606 3.4 11.7473 - L1 1636.606 8.3473 L1 u2 - 196.06 u = 114 m/s) Lett be the time of flight From a kinem ati c relation 1.6014 sin40 t49t2 1.60125.986 t -4.9t → 4.9 t2-125.986 t-1.60 = 0 On solving the qudratic equati on, t-25.724 s Horizontal di stance covered beyond the wall is X-u cos8 t - 14 - (14) cos40° x 25.724 - 14 - 261.88 m

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