Resolve the velocities into vertical (-v0sin(23°) and horizontal (v0cos(23°)) components.
The height of the building can be found from the vertical component.
The horizontal distance is the product of the time (6 seconds) and the horizontal component.
(c) can be found also from the vertical component:
10 = v0t+(1/2)(-g)t²
Solve for t.
A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 7.70 m/s at an angle of 23.0° below the horizontal. It strikes the ground 6.00 s later. (Ignore air resistance.)
(a) How far horizontally from the base of the building does the ball strike the ground?
(b) Find the height from which the ball was thrown.
(c) How long does it take the ball to reach a point 10.0 m below the level of launching?
1 answer
1 answer