Ball is thrown up with a velocity of 50 m/s.find the displacement and the distance traveled in 7 s.? Draw the motion on velocity time graph for first 10 s.

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Oddman answered
The displacement (d) at time (t) is given by
  d = v0*t - (1/2)g*t^2
where v0 is the initial velocity and g is the acceleration due to gravity, 9.8 m/s^2. After 7 seconds, the displacement will be
  d = 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 = v0/g = 50/9.8 ≈ 5.1 seconds
At that time, the ball has a height in meters of
  d = v0*(v0/g) - (1/2)g*(v0/g)^2
  = v0^2/(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 seconds
  127.55 m + (127.55 m - 109.9 m) = 145.2 m
A Velocity vs. Time curve can be found here.

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