1. a x = ut + 0.5at2
then 4.9 m = 0 + 0.5(9.8 m s–2)t2
and t = 1.0 s
b x = (average speed)(time) = (20 m s–1)(1.0 s) = 20 m
c The acceleration of the ball is constant at any time during its flight, and is equal to the acceleration due to gravity
= 9.8 m s–2 down
d After 0.80 s, the ball has two components of velocity:
vx = 20 m s–1
and vy = 0 + (9.8 m s–2)(0.80 s) = 7.84 m s–1
The speed of the ball at 0.80 s is given by:
[(20 m s–1)2 + (7.84 m s–1)2]½ = 21.5 m s–1
e The ball will strike the ground 1.0 s after it is struck.
Then vx = 20 m s–1
and vy = 0 + (9.8 m s–2)(1.0 s) = 9.8 m s–1
The speed of the ball at 1.0 s is given by:
[(20 m s–1)2 + (9.8 m s–1)2] ½ = 22.3 m s–1