A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. How long is the ball in flight before it hits the ground?

a. 9.14 s

b. 6.79 s

c. 2.44 s

d. 1.22 s

e. 7.45 s

Question

A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. How long is the ball in flight before it hits the ground?

a. 9.14 s

b. 6.79 s

c. 2.44 s

d. 1.22 s

e. 7.45 s

anonymous · Answer

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Break down the velocity into its vertical component, using trigonometry. $$sin \theta = \frac{opposite}{hypotenuse}$$You're looking for the 'opposite' side.

Then you can simply use $$v=u+at$$, where v is the final velocity of the ball (it will be zero at the point it hits the ground), u is the initial velocity (which you just worked out), a is the acceleration due to gravity (9.81 ms$^{-2}$), and solve for t!