A stone thrown from the top of a building is given an initial velocity of 20 m/s

Question

A stone thrown from the top of a building is given an initial velocity of 20 m/s straight upward. The building is 50 m high and the stone just misses the edge of the roof on its way down. Let g = 10.0 m/s2. a) Determine the time at which the stone reaches its maximum height b) the maximum height at B c) the time at which the stone returns to the height from which it was thrown d) the velocity of the stone at this instant at C e) the velocity and position of the stone at t = 5 s f) the time at the bottom g) the impact velocity on the ground.

Explanation / Answer

1.V= u+at 0= 20-10t t= 2sec 2.S= ut +1/2 at^2= 40+1/2 *-10*4= 20 m 3. s= 1/2 at^2 as the velocity at the top point is zero 20 = 1/2 *10*t^2 t=2sec 4. v^2 - u^2 = 2as v^2 = 2*10 * 20 v= 20m/s 5. v= u +at v= 20 +10 *5= 70m/s 6. s= 1/2 at^2 = 1/2 *10*t^2 70 = 1/2 *10*t^2= 3.74sec 7. v^2= 2*10*70= 1400 v= 37.4m/s

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