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a boy throws a ball upwards with an angle of 45 degrees of velocity 100 m/s that makes a parabola in the sky. what will be the Time taken by the ball, total distance and displacement?????
T = 2Usin@/g = 2*100/10sqrt2 =10sqrt2 (total time) displacement = range of projectile = u2sin2@/g=u2sin90/g =1000m
T = 2Usin@/g = 2*100/10sqrt2
=10sqrt2 (total time)
displacement = range of projectile = u2sin2@/g=u2sin90/g
=1000m
the initial velocity of the ball is 100m/s and angle of projection is 45 degrees therefore time taken by the ball to come to the ground is T=u*sin theta/2g the total distance covered by the ball and the displacement of the ball is the same thing the formula for the total distance covered by the ball is=horizontal speed*time taken by the ball horizontal speed of the ball is u*cos theta put the values and get the answer
the initial velocity of the ball is 100m/s and angle of projection is 45 degrees
therefore time taken by the ball to come to the ground is T=u*sin theta/2g
the total distance covered by the ball and the displacement of the ball is the same thing
the formula for the total distance covered by the ball is=horizontal speed*time taken by the ball
horizontal speed of the ball is u*cos theta
put the values and get the answer
The time taken by the ball is 2usin45/g.The displacement is the range of the projectile which comes out to be 1km.
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