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A ball is thrown in vertically upwards direction with velocity 50 m/s. Maximum height covered by the ball will be ………… (g = 10 m/s2) |
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Answer» tion:(a) INITIAL velocity of ball (u)=50m/s, ACCELERATION of ball =−g,FINAL velocity at the highest point (v)=0So applying the 3rd equation of motion we get:v 2 =u 2 −2gh max ⇒0=50 2 −2×10×h max ⇒h max = 202500 =125m(B) Let the time required to reach max height be t. Then applying 1st equation of motion we get:v=u−gt⇒0=50−10t⇒t=5s(c) Let SPEED at half of max height be V then:V 2 =50 2 −2g 2125 ⇒V 2 =2500−1250=1250⇒V= 1250 =35.35m/s. |
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