cricket ball is hit vertically
upwards and returns to ground
6 s later. Calculate (i) maximum
height reached by the ball,
(ii) initial velocity of the ball
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A cricket ball is hit vertically upwards and returns to the ground after 6 seconds. We need to find the maximum height reached and the initial velocity. This is a classic projectile motion problem under gravity.
Let's analyze the given information. The ball takes 6 seconds total to go up and come back down. Due to the symmetry of vertical motion under gravity, the time to reach maximum height is exactly half the total time, which is 3 seconds.
To find the initial velocity, we use the kinematic equation v equals u plus at. At maximum height, the velocity is zero. The time to reach maximum height is 3 seconds, and acceleration due to gravity is negative 9.8 meters per second squared. Substituting these values, we get u equals 29.4 meters per second.
To find the maximum height, we use another kinematic equation: v squared equals u squared plus 2as. At maximum height, velocity is zero. We know the initial velocity is 29.4 meters per second and acceleration is negative 9.8. Solving this equation gives us a maximum height of 44.1 meters.