A particle projected vertically upwards from a point A on the ground. It takes t1 sec to reach appoint B at a height “h” from B but still continues to move up. If it further takes t2 sec from B to the ground again, then find maximum height reached.
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Consider the ball is thrown from the tower to the ground in vertical manner. So, the initial velocity = 0m/s.
Time taken = t sec and distance travelled= h meters. Acceleration = 9.8 ms^-2.
The equation of motion S=ut + ½ at^2. Then H1/H2 =(t1/t2)^2. We solve and obtain H2 = H1/9 from the tower top.
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