A ball is tossed from an upper-story window of a building. The ball is given an

Question

A ball is tossed from an upper-story window of a building. The ball is given an initial velocity of 20m/s at an angle of 37 degree below the horizontal. It strikes the ground 3 seconds later.
a).- Determine the horizontal and vertical components of the initial velocity.
b).- How far horizontally from the base of the building does the ball strikes the ground?
c).- Find the height from which the ball was thrown
d).- how long does it take the ball to reach a point 25 m bellow the level of launching?
Can anybody help me with this problem please? I really appreciate all your answers
Thank you.

Explanation / Answer

Hi,   Given u = 20 m/s and angle ? with +ve x-axis = 37 deg (below). a) Hence horizontal velocity = ux = u * Cos(?) = 15.973 m/s       and vertical velocity = uy = u * Sin(?) = 12.0364 m/s b) The ball strikes the ground in 3 seconds. And there is no other force acting in horizontal direction. Hence the horizontal velocity is constant and hence the horizontal distance travelled = x = ux * t = 15.973 * 3 = 47.919 m c) Along with the initial vertical velocity, the ball is also acted on by the vertical gravitational pull with an acceleration of g. Hence from s = ut + (1/2)at^2 we have,       s = 12.0364(3) + (0.5 * 9.8 * 3^2) = 80.2092m. Hence its thrown from a height of 80.21m (app.). d) for travelling a distance of 25m the time taken can be found from the above equation again as below: 25 = 12.0364(t) + (0.5 * 9.8 * t^2) on simplifying we get, t = 1.343sec d) for travelling a distance of 25m the time taken can be found from the above equation again as below: 25 = 12.0364(t) + (0.5 * 9.8 * t^2) on simplifying we get, t = 1.343sec Hope this helps you.

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