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Example 1. An iron ball of mass 3 kg is released from a height of 125 m and fallsfreely to the ground. Assuming that thevalue of g is 10 m/s?, calculate(i) time taken by the ball to reach theground(ii) velocity of the ball on reaching theground(iii) the height of the ball at half the time ittakes to reach the ground.Given: m = 3 kg, distance travelled by theball s = 125 m, initial velocity of the ball =u=o and acceleration a = g = 10 m/s2.(i) Newton's second equation of motiongives |
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Answer» T+1/2AT^2as U is =0therefore S=1/2AT^2here A is 10m/s^2therefore 125=1/2AT^2250=10T^2therefore T =√25=5secV^2-U^2=2AShere A is -10 and S is -125m as its FALLING from a height and U =0 thereforeV^2=2(-10)(-125)V=√2500V=50m/shalf time by it reaches =5/2=2.5secS=UT+1/2AT^2S=1/2x10xT^2therefore S=31.25mbut H=125-31.25therefore H=93.75ma)The time by which the ball reaches the ground is=5secondsb) The VELOCITY of the ball on reaching the ground=50m/sc) The height of the ball at half time it TAKES to REACH the ground=93.75mI hope this will HELP you |
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