Question

A man stands on the roof of a building of height 16.3 m and throws a rock with a velocity of magnitude 25.8 m/s at an angle of 27.6 ∘ above the horizontal. You can ignore air resistance. Calculate the magnitude of the velocity of the rock just before it strikes the ground. Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.

Answer 1

Use the conservation of energy of rock at initial position (roof) and at final position (at impact) to find the velocity at impact. KE; +PE; = KE, +PE, 3 mu? + mgh = = mv +0 | < +2gh=v} +0 v; = \u? +2gh = V(25.8 m/s)* +2(9.81)(16.3 m) = 31.3918 m/s 31.4 m/s Therefore, the magnitude of the velocity of the rock just before it strikes the ground is 31.4 m/s --------------------------------------------------------- Use the formula hmas = (usin o) to find the maximum height above the roof reached by the rock if required. 2g

First find the time to reach the ground and then later find the horizontal distance. Sul ut + -at2 take the vertical distance and vertical velocity from the roof. (-16.3 m)= (usin )t + }(-9.81 m/s?)?? (-16.3 m)=(usi (-16.3 m) = (25.8 m/s )sin 27.6°)1+2(–9.81 m/s?)t? 4.905+ –11.951–16.3=0 11.951. (11.952) – 4(4.905)(-16.3) t=- 9.81 t=3.41 s Horizontal distance (range) of the rock is, R=(u cost =((25.8 m/s)cos 27.6°)(3.41 s) = 77.96 m Therefore, horizontal distance from the base of the building to the point where the rock strikes the ground is 77.96 m