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A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball? |
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Answer» Let (u) be the velocity of projection of the ball. The ball will cover maximum horizontal distance when angle or projection with whorizontal, ` theta = 45^@. Then, ` R_(max) = u^2 //g` …(i0 In order to study to the motion of the ball along vertcal direction, consider a point on the surface of earth as the origin and vertical upward direction as the positive direction of (Y) axis . Taking motion of the ball along vertical upward direction we have ` u_y = u, a_y =- g. v_y =0, t= ?, y+0 =0 y=?` Also , ` y=y_0 = u_y t = 1/2 a_y t^2` :. ` y=0 + u ( u /g) + 1/2 (- g ) u^2 /g - 1/2 u^2 /g = 1/2 = ( 1000/2 = 50 m` [ from (i) ]` . |
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