During a baseball game, a batter hits a high pop up. If the ball remains in the air for 4.16 S, how high above the point where it hits the bat does it rise? Assume when it hits the ground it hits at exactly the level of the bat. The acceleration of gravity is 9.8 M/S2. Answer in units of m
To determine the height the ball rises above the point where it hits the bat, we can use the equations of motion in projectile motion. We know that the ball is initially at rest when it leaves the bat and the only force acting on it is gravity. We also know the total time of flight, which is 4.16 seconds, and the acceleration due to gravity, which is 9.8 m/s^2.
First, let's find the time it takes for the ball to reach the highest point of its trajectory. In projectile motion, the time taken to reach the highest point is equal to half of the total time of flight. So, the time taken to reach the highest point is 4.16 seconds divided by 2, which gives us 2.08 seconds.
Next, we can use the equation of motion for vertical displacement to calculate the height the ball rises above the point where it hits the bat. The equation is:
h = Vāy * t + (1/2) * a * t^2
Where:
h = vertical displacement or height
Vāy = initial vertical velocity (which is 0 since the ball starts from rest)
t = time
a = acceleration due to gravity (which is -9.8 m/s^2 since gravity acts downwards)
Since the initial vertical velocity is 0 and the acceleration due to gravity is negative, the equation simplifies to:
h = (1/2) * a * t^2
Plugging in the values, we get:
h = (1/2) * (-9.8 m/s^2) * (2.08 s)^2
Calculating this expression gives us:
h = -9.8 m/s^2 * 4.3264 s^2
h ā -42.40 m^2/s^2
Since we are interested in the height above the point where the ball hits the bat, the negative sign can be ignored in this context.
Therefore, the ball rises approximately 42.40 meters above the point where it hits the bat.
The initial v = 9.81*2.08 = 20.40 m/s (why?)
so, the height is
h = 40.8*2.0 - 4.9*2.08^2 = 60.4 m