Question #253389

A ball on the ground is shot straight up and it reaches a maximum height of 25 m/s. (a) What was the ball’s initial velocity? (b) After reaching its maximum height, how long will the ball take before it lands again? (c) What is the ball’s speed just before it hits the ground?


Expert's answer

Given:

hmax⁡=25 mh_{\max}=25\:\rm m

g=9.8 m/s2g=9.8\:\rm m/s^2



(a) the ball’s initial velocity

v0=2ghmax⁡=2∗9.8∗25=22 m/sv_0=\sqrt{2gh_{\max}}=\sqrt{2*9.8*25}=22\:\rm m/s

(b) the time of free faling

t=2hmax⁡g=2∗259.8=2.3 st=\sqrt{\frac{2h_{\max}}{g}}=\sqrt{\frac{2*25}{9.8}}=2.3\:\rm s

(c) the final speed

vf=v0=22 m/sv_f=v_0=22\:\rm m/s


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