A particle is projected vertically up from the top of a tower with velocity 10m/s

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u=10m/s t=5 s Using 2nd law of motion , i.e. s=ut+1/2at^2 Here a=g=-9.8m/s^2 =10×5-1/2×9.8×5×5 =50-122.5 =72.5m

A particle is projected vertically up from the top of a tower with velocity 10 m/sec. It reaches the ground in 5 sec. Find, (i) Height of the tower, (ii) Striking velocity of particle at ground, (iii) Distance traversed by the particle and Average speed and average velocity of particle?

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