A lady, standing on level horizontal ground, throws a ball. When it leaves her h

Question

A lady, standing on level horizontal ground, throws a ball. When it leaves her hand it is 2 m above the ground, its velocity is 25 ms?1 and it is travelling at an angle of 40? above the horizontal. Assume that the ball is a particle and that its weight is the only force that acts on the ball after it has left her hand.

Let A be the point where it leaves her hand, H be the point of maximum height and B be the point where it lands on the ground, as shown.

(a) Calculate the horizontal and vertical components of velocity at A.

(b) Find the time of flight from A to B.
(c) Find the horizontal distance that it travels.

(d) Find the maximum height of the ball above the ground.

Explanation / Answer

a)horizontal component = 25 cos 40 = 19.15 m/s

vertical component = 25 sin 40 = 16.07 m /s

d) in vertical direction at highest point

v^2 = u^2 - 2gh

0 = 16.07^2 - 2*9.8*h

h = 13.175 m

max height of ball above ground = 15.175 m

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