Question

A brick is thrown upward from the top of a building at an angle of 40

Answer 1

The horizontal component of this velocity = 17 * cos(40) = 13.02 m/s and is constant.

The vertical component = 17 * sin(40) = 10.93 m/s and changes because of gravity.

First find how long it takes an object to reach its maximum height when its initial upward speed is 10.93 m/s. This is the same time it would take for any dropped object to go from zero to 10.93 m/s.

V = g *t, so t = v / g = 10.93 / 9.8 = 1.11 seconds.

D = 1/2 g t^2 = 4.9 * 1.11^2 = 6.0 meters.

The brick has been moving for 1.11 seconds, so there are 2 seconds left before it hits the ground. In two seconds it will fall a distance of 4.9 * 2^2 = 20 meters, so the height of the building is 20-6 = 14 meters.