A boy tosses a football upward. If the football rises a vertical distance of 1.1 m and the boy catches it at the same point he released it, what is the velocity of the ball just before he catches it?
V^2 = 2 g H
where H is the height it rises, g is the acceleration of gravity and V is the velocity when thrown.
V = 4.643 m/s
When caught, it has the same speed but opposite direction (down).
A boy tosses a football upward. If the football rises a vertical distance of 3.7 m and the boy catches it at the same point he released it, what is the velocity of the ball just before he catches it?
To find the velocity of the ball just before the boy catches it, we need to use the equation of motion for objects in free fall:
Vf = Vi + (a * t)
Here,
Vf = final velocity of the ball
Vi = initial velocity of the ball
a = acceleration due to gravity (approximately -9.8 m/s^2, considering downward direction as negative)
t = time taken for the ball to reach the highest point and then fall back down
Since the ball rises to a height of 1.1 m before falling back down to the same position, the total height is 2 * 1.1 = 2.2 m.
Using the equation for height in free fall:
2.2 = (Vi * t) + (0.5 * a * t^2)
We know that the final velocity at the highest point will be 0 m/s, so we can set:
Vf = 0
Therefore, the equation becomes:
0 = Vi + (a * t)
Now, let's solve the equation for Vi:
Vi = - (a * t)
Substituting the value of acceleration due to gravity, a = -9.8 m/s^2:
Vi = 9.8 * t
By using the equation for height in free fall:
2.2 = (9.8 * t * t) / 2
2.2 = 4.9 * t^2
Simplifying:
t^2 = 2.2 / 4.9
t^2 = 0.44898
Taking the square root of both sides:
t = √0.44898
t ≈ 0.67 s
Now, substituting this value of t into the equation for initial velocity:
Vi = 9.8 * 0.67
Vi ≈ 6.56 m/s
Therefore, the velocity of the ball just before the boy catches it is approximately 6.56 m/s.
To find the velocity of the ball just before the boy catches it, we can use the equation for the final velocity of an object undergoing uniform acceleration:
v^2 = u^2 + 2as
where:
- v is the final velocity (which we want to find)
- u is the initial velocity (the velocity at which the boy released the ball)
- a is the acceleration due to gravity (which is approximately -9.8 m/s^2 downwards, since we're dealing with upward motion)
First, we need to find the value of u. Since the boy catches the ball at the same point he released it, the displacement (s) is equal to zero. Therefore, the equation becomes:
v^2 = u^2 + 2(0)(0)
Simplifying this equation, we get:
v^2 = u^2
Taking the square root of both sides, we have:
v = u
This means that the final velocity of the ball just before the boy catches it is equal to the initial velocity at which the boy released it.