Question

A ball is launched directly upward from ground level with an initial speed of 18 m/s. (Air resistance is negligible.) (a) How long is the ball in the air? 3.67 (b) What is the greatest height reached by the balil? 16.53 (c) How many seconds after launch is the ball 7 m above the release point? 3.23 x s after being thrown (first time) x s after being thrown (second time) 0.44 Be careful when assigning signs to the kinematic quantities. What is the speed of the ball when it is at its greatest height?

Answer 1

a)

From

Y=Y_o+V_ot-(1/2)gt²

when ball hits ground Y=0

0=0+18t-(1/2)*9.8*t²

t=18*2/9.81 =3.67 s

b)

From

V²=V_o²-2gH

at maximum height V=0

0=18²-2*9.8*H

H=18²/19.62 =16.53 m

c)

at Y=3

=>7==0+18t-(1/2)*9.8*t²

4.9t²-18t+7=0

t₁=0.44 s (first time)

t₂=3.23 s (second time)

Enter answer 0.44 s for first time

and 3.23 s for second time