A man can throw a ball a maximum horizontal
distance of 142 m.
The acceleration of gravity is 9.8 m/s
2
.
How far can he throw the same ball vertically upward with the same initial speed?
Answer in units of m.
To find out how far the man can throw the ball vertically upward with the same initial speed, we can use the equations of motion.
When the ball is thrown vertically upward, the only force acting on it is gravity, which causes a downward acceleration. The initial velocity is the same as when the ball was thrown horizontally, but it acts in the opposite direction.
The equation to find the maximum height reached by an object thrown vertically upward is given by:
h = (v^2) / (2g)
Where:
h is the maximum height,
v is the initial velocity, and
g is the acceleration due to gravity.
In this case, we don't know the initial velocity. However, we know the initial horizontal velocity, and we can use it to find the initial vertical velocity.
Since the horizontal and vertical motions are independent, the initial horizontal velocity does not affect the initial vertical velocity. Therefore, we can assume the initial vertical velocity is the same as the initial horizontal velocity. Let's call this initial velocity "u."
Now, to find the initial vertical velocity, we can use the following equation:
u = v * sinθ,
Where:
u is the initial vertical velocity,
v is the initial velocity (in this case, the horizontal velocity), and
θ is the angle of projection.
Since the ball is thrown straight up, θ is 90 degrees. Hence, sinθ is equal to 1.
So, the equation becomes:
u = v * sin90,
u = v * 1,
u = v.
This means that the initial vertical velocity is equal to the initial velocity (horizontal velocity).
Now, we can substitute the value of u into the equation for the maximum height:
h = (u^2) / (2g).
Since u = v, we can write:
h = (v^2) / (2g).
We are looking for the maximum height, and the ball will reach its highest point when it comes to a stop momentarily before falling back down. At this point, the vertical velocity becomes zero.
So, we can find the maximum height by calculating the vertical distance traveled when the vertical velocity is zero. We can use the equation:
v = u - gt,
Where:
v is the final vertical velocity (which is zero),
u is the initial vertical velocity (which is equal to v),
g is the acceleration due to gravity, and
t is the time taken.
Rearranging the equation, we can solve for t:
t = u / g.
Now, we can substitute the value of t into the equation for the maximum height:
h = (v^2) / (2g)
h = (u^2) / (2g)
h = (v^2) / (2g)
h = (v^2) / (2g)
Substituting the value of t into the equation:
h = (v^2) / (2g)
h = (u^2) / (2g).
Now, we can substitute the known values into the equation. The acceleration due to gravity is 9.8 m/s^2, and we know that the horizontal velocity is 142 m/s.
h = (142^2) / (2 * 9.8),
h = 20164 / 19.6,
h = 1028.57 m.
Therefore, the man can throw the same ball vertically upward with the same initial speed for a maximum height of approximately 1028.57 meters.