A man stands on the roof of a 19.0{\ m m} -tall building and throws a rock with

Question

A man stands on the roof of a 19.0{ m m} -tall building and throws a rock with a velocity of magnitude 24.0{ m m}/{ m s} at an angle of 42.0 degrees above the horizontal. You can ignore air resistance.. Part A Calculate the maximum height above the roof reached by the rock. I know h max is 13.2. Part B Calculate the magnitude of the velocity of the rock just before it strikes the ground. Part C Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Explanation / Answer

The vertical component of velocity when the rock is thrown is 30 sin 42 = 20.07 m/s Calculate the time T that it takes the rock to hit the ground using that initial velocity. Hint: 20.07 T - (g/2) T^2 + 19.0 = 0 I assume you know what g is, and how to solve that quadratic equation. Take the positive root. T times the horizontal velocity component is the horizontal distance that the rock travels. The maximum height h of the rock is attained when the vertical velocity component is zero. This happens when gt = 20.07 m/s h = (1/2)(20.07 m/s)^2/g

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