Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
Consider a bicycle that has wheels with a circumference of 2 m. Solve for the linear speed of the bicycle when it's wheels rotat
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Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
The acceleration of the ball after leaving the hand is downward
Explanation:
In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.
After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is
where m is the mass of the ball and g is the acceleration of gravity (), acting in the downward direction.
According to Newton's second law, the acceleration of the ball is given by
where
is the net force acting on the ball
After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:
This means that the acceleration of the ball remains downward for the entire motion, after leaving the hand.
Learn more about Newton's second law:
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