A boy throws a ball so that it may just clear a wall 3.6m high. the boy is at distance of 4.8m from the wall the ball was found to hit the ground at a distance of 3.6m on the other side of the wall. find the least velocity with the ball can be thrown
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x=x0+v0*cos(o)*t , where x is the final distance, x0 the initial position, v the velocity you're looking for , o the angle and t the time.
y=y0+v0*sin(o)*t-.5*g*t² , where y is the final height, y0 the initial height and g the acceleration of gravity.
v=v0+g*t , where v is the current velocity and v0 the initial velocity.
You know that x0 and y0 are equal to 0, so the problem is simplified.
You want to know the velocity for which y=3.6, x=4.8/2=2.4 (since it's at his peak) and v = 0.
Now you have three equations and three incognites, therefore you can solve it easily.
y=y0+v0*sin(o)*t-.5*g*t² , where y is the final height, y0 the initial height and g the acceleration of gravity.
v=v0+g*t , where v is the current velocity and v0 the initial velocity.
You know that x0 and y0 are equal to 0, so the problem is simplified.
You want to know the velocity for which y=3.6, x=4.8/2=2.4 (since it's at his peak) and v = 0.
Now you have three equations and three incognites, therefore you can solve it easily.
ArunMani1:
no brother answer is u=9.7745 ...but to solve
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