A home run just clears a fence 105 m from home plate. The fence is 4.00 m higher than the height at which the batter struck the ball, and the ball left the bat at a 31.0° angle above the horizontal. At what speed did the ball leave the bat
To find the speed at which the ball left the bat, we can use the principles of projectile motion.
First, let's break down the problem into its components. We have the horizontal and vertical motion of the ball. The horizontal motion is unaffected by gravity, while the vertical motion is influenced by the force of gravity.
We are given the following information:
- The horizontal distance the ball traveled, which is 105 m.
- The vertical height difference between the fence and the height at which the batter struck the ball, which is 4.00 m.
- The angle at which the ball was hit above the horizontal, which is 31.0°.
Now, let's analyze the vertical motion. The ball starts at a certain height and returns to the same height when landing over the fence. The only forces acting on the ball vertically are gravity and the initial vertical velocity.
Using the kinematic equation for vertical motion:
(y = yo + voy*t - (1/2)*g*t^2),
where:
- y is the displacement in the vertical direction (4.00 m),
- yo is the initial height (0 m),
- voy is the initial vertical velocity (unknown),
- g is the acceleration due to gravity (approximately 9.8 m/s²),
- t is the time.
Since the vertical displacement is symmetrical (the ball reaches the same height as it started), we can find the time for half the total displacement.
So, using the equation:
(y = yo + voy*t - (1/2)*g*t^2),
and substituting for the values of y (4.00 m) and yo (0 m), and rearranging the equation:
4.00 m = voy*t - (1/2)*(9.8 m/s²)*t^2.
Now, let's analyze the horizontal motion. The horizontal distance traveled by the ball is 105 m, and there are no horizontal forces acting on the ball (assuming no air resistance).
The horizontal distance (x) is related to the time (t) and initial horizontal velocity (vox) using the equation:
x = vox*t.
Now, we have the values of x (105 m) and the angle (31.0°). We can relate the initial horizontal velocity and initial vertical velocity using the angle.
The initial horizontal velocity (vox) can be found using trigonometry:
vox = v*cos(theta),
where v is the initial velocity of the ball and theta is the angle above the horizontal (31.0°).
Now, let's connect the horizontal and vertical motion. We can rearrange the equation for horizontal motion (x = vox*t) to solve for t:
t = x/vox.
Substituting the value of t into the equation for vertical motion (4.00 m = voy*t - (1/2)*(9.8 m/s²)*t^2) gives us:
4.00 m = voy*(x/vox) - (1/2)*(9.8 m/s²)*(x/vox)^2.
Now, we have an equation with two unknowns: voy (initial vertical velocity) and vox (initial horizontal velocity). Since vx = vox (horizontal velocity is constant), we can rewrite the equation as:
4.00 m = voy*(x/vx) - (1/2)*(9.8 m/s²)*(x/vx)^2.
Now, we substitute the known values for x and the angle into the equation:
4.00 m = voy*(105 m)/(v*cos(31.0°)) - (1/2)*(9.8 m/s²)*(105 m/(v*cos(31.0°)))^2.
We can solve this equation to find the value of voy (initial vertical velocity). Rearranging the equation and isolating voy gives us:
voy = (4.00 m)*(v*cos(31.0°))/(105 m + (1/2)*(9.8 m/s²)*(105 m/(v*cos(31.0°)))^2).
Finally, to find the initial speed (v) at which the ball left the bat, we substitute the value we obtained for voy into the equation:
v = sqrt((voy^2 + vox^2)),
where vox = v*cos(31.0°).
Using this equation, we can calculate the speed at which the ball left the bat.