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 42deg above the horizontal, giving it an initial velocity of 15m/s. If the ball is 5.3m 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?
Use energy:
Initial KE+initialPE=finalKE
1/2 m 15^2 + m*g*5.3= 1/2 m vf^2
To solve this problem, we can use basic kinematic equations for motion in two dimensions. Let's break down the given information:
1. Initial vertical velocity (Vy0) = 15 m/s (sin 42°)
- The initial velocity has a vertical component calculated by multiplying the initial velocity (15 m/s) with the sine of the launch angle (42°).
2. Initial horizontal velocity (Vx0) = 15 m/s (cos 42°)
- The initial velocity also has a horizontal component calculated by multiplying the initial velocity (15 m/s) with the cosine of the launch angle (42°).
3. Initial vertical position (h0) = 5.3 m
- The ball starts 5.3 meters above the level of the playing field.
4. Final vertical position (hf) = 0 m
- The ball will hit the ground, so the final vertical position is 0 meters.
Now, using the equation for vertical motion:
hf = h0 + Vy0t - (1/2)gt^2
Considering that hf = 0, h0 = 5.3, and g (acceleration due to gravity) = -9.8 m/s^2 (as it acts in the opposite direction of the motion), we can calculate the time of flight (t).
0 = 5.3 + (15 sin 42°)t - (1/2)(-9.8)t^2
The resulting quadratic equation can be solved for t, which will give us the time it takes for the ball to hit the ground.
After finding t, we can use the following equation to find the final horizontal position (x):
x = Vx0 * t
Since the final vertical position is 0, we've already determined that the vertical component of velocity (Vy) will be 0 m/s at that point. We can use the Pythagorean theorem to find the magnitude of the final velocity (Vf).
Vf^2 = (Vx)^2 + (Vy)^2
Since Vy = 0 m/s, we are left with:
Vf = Vx