A tennis ball is thrown horizontally from an elevation of 20.60 m above the ground with a speed of 21.0 m/s
vertical m from ground???
f the ball is still in the air, how long before it hits the ground? (If the ball is not in the air, enter 0.)????
If the ball is still in the air, where will it be with respect to the starting point once it lands???
To calculate the vertical position and time it takes for the tennis ball to hit the ground, we can use the equations of motion.
1. Vertical distance from the ground:
Since the ball is thrown horizontally, there is no initial vertical velocity. Therefore, the vertical position of the ball can be calculated using the equation:
s = ut + (1/2)gt^2
where s is the vertical distance, u is the initial velocity in the vertical direction (0 m/s in this case), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Given: u = 0 m/s, g = 9.8 m/s^2, and s = ? (to be determined)
Using the equation, we can rearrange it to solve for s:
s = (1/2)gt^2
s = (1/2) * 9.8 * t^2
s = 4.9t^2 (equation 1)
Substituting the given values, we can solve for s.
2. Time before the ball hits the ground:
The time it takes for the ball to hit the ground can be found by considering the vertical motion. The ball starts from an initial height and falls under the influence of gravity until it reaches the ground.
The equation for the time of flight can be used:
s = ut + (1/2)gt^2
where s is the vertical distance (which is now 20.60 m), u is the initial velocity, g is the acceleration due to gravity, and t is the time.
Given: s = 20.60 m, u = 0 m/s, g = 9.8 m/s^2, and t = ? (to be determined)
Using the equation, we can rearrange it to solve for t:
s = (1/2)gt^2
20.60 = (1/2) * 9.8 * t^2
41.2 = 4.9t^2
t^2 = 41.2 / 4.9
t^2 = 8.41
t ≈ √8.41
t ≈ 2.9 seconds (equation 2)
3. Horizontal position:
Since the ball was thrown horizontally, its horizontal velocity remains constant throughout the motion. Therefore, the horizontal position is determined by multiplying the horizontal velocity by the time of flight.
Given: Horizontal velocity = 21.0 m/s
Time = 2.9 seconds (from equation 2)
Horizontal position = Horizontal velocity * Time
Horizontal position = 21.0 m/s * 2.9 s
Horizontal position ≈ 60.9 m
Therefore, the tennis ball lands approximately 60.9 m horizontally away from the starting point.
To summarize:
- The vertical distance from the ground is given by equation 1: s = 4.9t^2
- The time it takes for the ball to hit the ground is given by equation 2: t ≈ √(s/4.9)
- The horizontal position where the ball lands is given by multiplying the horizontal velocity by the time of flight.