The displacement (d) at time (t) is given by

d = v

where v

d = 50*7 - (1/2)*9.8*7^2

= 350 - 4.9*49

= 350 - 240.1

= 109.9

The

The time at which the ball reaches its peak height is

t = v

At that time, the ball has a height in meters of

d = v

= v

= 50^2/19.6

≈ 127.55

Its total

127.55 m + (127.55 m - 109.9 m) =

A Velocity vs. Time curve can be found here.

d = v

_{0}*t - (1/2)g*t^2where v

_{0}is the initial velocity and g is the acceleration due to gravity, 9.8 m/s^2. After 7 seconds, the displacement will bed = 50*7 - (1/2)*9.8*7^2

= 350 - 4.9*49

= 350 - 240.1

= 109.9

The

**displacement**after 7 seconds is**109.9 meters.**The time at which the ball reaches its peak height is

t = v

_{0}/g = 50/9.8 ≈ 5.1 secondsAt that time, the ball has a height in meters of

d = v

_{0}*(v_{0}/g) - (1/2)g*(v_{0}/g)^2= v

_{0}^2/(2g)= 50^2/19.6

≈ 127.55

Its total

**distance traveled**is then the height to the peak plus the distance back down to where the ball is at 7 seconds127.55 m + (127.55 m - 109.9 m) =

**145.2 m**A Velocity vs. Time curve can be found here.