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A particle is projected with velocity \(v_0\) horizontally from a height \(H\) over a hill, and strikes the hill perpendicularly.

Find \(2gH/v_0 ^ 2\)

Assume the hill rises linearly from the origin at an angle \(\theta = \dfrac\pi6 = \si{30}^\circ\).

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Which of the following statements are true for a basketball that is projected diagonally upwards (air resistance **cannot be ignored**)?

a. The basketball's kinetic energy is zero at the highest point of its motion.

b. The basketballâ€™s average kinetic energy is smaller when traveling upwards than downwards.

c. The time taken for the basketball to travel upwards is less than the time taken to travel downward.

d. The horizontal distance traveled when the basketball is moving upwards is equal to the horizontal distance traveled when the basketball is moving downwards.

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A projectile is launched from an initial position \((x,y) = (-\SI{4}{\meter},\SI{0}{\meter})\). The projectile's initial speed is \(12\text{ m/s}\) and it is launched at a \(\SI{75}{\degree}\) angle with respect to the \(x\)-axis. There is a staircase with 4 steps, as pictured above. The top step begins at \((x,y) = (\SI{0}{\meter},\SI{4}{\meter})\) meters and the bottom step ends at \((x,y) = (\SI{4}{\meter},\SI{0}{\meter})\). Each step is \(\SI{1}{\meter}\) wide and \(\SI{1}{\meter}\) deep. The projectile flies over part of the staircase and lands on one of the steps. What is the \(x\) coordinate of the landing point in meters?

**Details and Assumptions**

- The gravitational acceleration is in the \(-y\) direction, with a magnitude of \(\SI{9.8}{\meter\per\second\squared}\).

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What should be the height \(H\) in meters so that the ball of mass \(\SI{1}{\kilo\gram}\) reaches the point \(A?\)

Take \(g=\SI[per-mode=symbol]{10}{\meter\per\second\squared}.\)

- There is no friction in this system.
- The ball is not rolling it is sliding.

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