Darryl Strawberry hits a baseball with a horizontal velocity of 45 m/s and a vertical velocity of 25 m/s...
Find the time it takes for the ball to travel to the highest point
Find the total time the ball is in flight
Find the range
Find the max height the ball reaches
Does it clear a fence 2 meters high 100 meters away?
Thanks..
To find the time it takes for the ball to travel to the highest point, we need to consider the vertical velocity of the ball. The vertical velocity decreases due to the force of gravity until it reaches its highest point, where it becomes zero. We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the initial vertical velocity is 25 m/s, and the final vertical velocity at the highest point is zero (since the ball reaches its maximum height and starts falling back down). We know that the acceleration due to gravity is -9.8 m/s^2 (negative because it acts opposite to the direction of motion). Substitute these values into the formula:
0 = 25 + (-9.8)t
-25 = -9.8t
Solving for t, we find:
t = 2.55 seconds
So, it takes approximately 2.55 seconds for the ball to reach its highest point.
To find the total time the ball is in flight, we can consider the time it takes for the ball to reach its highest point and then return to the ground. Since the vertical motion is symmetrical (it takes the same amount of time to reach the highest point as it does to fall back down to the ground), we can double the time it takes for the ball to reach the highest point:
Total time in flight = 2 * 2.55 seconds = 5.1 seconds
The range of the ball can be determined by considering its horizontal velocity. The range is the horizontal distance traveled by the ball during the total time in flight. We can use the formula:
Range = horizontal velocity * total time in flight
Range = 45 m/s * 5.1 seconds = 229.5 meters
So, the range of the ball is approximately 229.5 meters.
To find the maximum height the ball reaches, we can use the formula h = ut + (1/2)at^2, where h is the height, u is the initial velocity, t is the time, and a is the acceleration. Since the initial vertical velocity is 25 m/s and the ball reaches its highest point where the final velocity is zero, we can substitute these values into the formula:
h = 25 * 2.55 + (1/2)(-9.8)(2.55)^2
h ≈ 32.12 meters
So, the ball reaches a maximum height of approximately 32.12 meters.
To determine if the ball clears a fence 2 meters high and 100 meters away, we need to consider the vertical and horizontal components of the ball's motion. The vertical velocity of the ball at any point during its flight can be calculated using the formula v = u + at. At the highest point, the vertical velocity is zero. We can calculate the time it takes for the ball to reach the height of the fence as follows:
0 = 25 + (-9.8)t
-25 = -9.8t
t ≈ 2.55 seconds
Now, we can calculate the vertical distance covered by the ball during this time:
Distance = ut + (1/2)at^2
Distance = 25 * 2.55 + (1/2)(-9.8)(2.55)^2 ≈ 32.12 meters
Since the height of the fence is 2 meters, we can see that the ball easily clears the fence.
To know if the ball clears the fence, we also need to check the horizontal distance traveled by the ball. The ball travels horizontally at a velocity of 45 m/s for a total time of 5.1 seconds. We can calculate the horizontal distance covered by the ball during this time:
Distance = velocity * time
Distance = 45 * 5.1 ≈ 229.5 meters
Since the horizontal distance traveled by the ball is greater than the distance to the fence (100 meters), we can conclude that the ball clears the 2-meter high fence.