During baseball practice, you go up into the bleachers to retrieve a ball. You throw the ball back into the playing field at an angle of 42° above the horizontal, giving it an initial velocity of 15 m/s. If the ball is 5.3 m above the level of the playing field when you throw it, what will be the velocity of the ball when it hits the ground of the playing field?

Question

During baseball practice, you go up into the bleachers to retrieve a ball. You throw the ball back into the playing field at an angle of 42° above the horizontal, giving it an initial velocity of 15 m/s. If the ball is 5.3 m above the level of the playing field when you throw it, what will be the velocity of the ball when it hits the ground of the playing field?

Answer 1

I assume you want final speed, not velocity.

final KE=Initial KE+initial PE
1/2 m vf^2=1/2 m(15)^2 +mg(5.3)
solve for Vf

Answer 2

To find the velocity of the ball when it hits the ground, you need to break down the motion of the ball into its vertical and horizontal components.

First, let's analyze the vertical component. We can use the equation of motion for vertical motion, which is given by:

y = yo + vot + (1/2)gt^2

where:
y = vertical position (in this case, it is the height above the playing field)
yo = initial vertical position (5.3 m)
vo = initial vertical velocity (unknown)
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight (unknown)

For the vertical component of the ball's motion, we know the following:
- The initial vertical position (yo) is 5.3 m.
- The initial vertical velocity (vo) can be found by multiplying the initial velocity (15 m/s) by the sine of the launch angle (42°).
- The acceleration due to gravity (g) is -9.8 m/s^2.

Let's calculate the initial vertical velocity (vo):
vo = 15 m/s * sin(42°)

Next, we can use the kinematic equation to find the time of flight (t) for the ball to hit the ground. The equation is:

y = yo + vot + (1/2)gt^2

Rearranging to solve for t:

0 = 5.3 m + (vo)t + (1/2)(-9.8 m/s^2)t^2

Simplify the equation to:
(1/2)(-9.8 m/s^2)t^2 + (vo)t + (5.3 m) = 0

Solve the quadratic equation using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = (1/2)(-9.8 m/s^2), b = vo (found from the previous step), and c = 5.3 m.

Once you have calculated the time of flight (t), you can use it to find the horizontal component of the ball's velocity. Multiply the initial horizontal velocity (15 m/s) by the cosine of the launch angle (42°).

Finally, you can combine the horizontal and vertical components of the ball's velocity using vector addition to find the total velocity just before it hits the ground.