A football is thrown with an initial upward velocity component of 13.5m/s and a horizontal velocity component of 16.5m/s

How much time is required for the football to reach the highest point in its trajectory? How high does it get above the ground?How much time after it is thrown does it take to return to its original height? How far has the football traveled horizontally from its original position?

When the ball reaches the highest point, it means that its vertical velocty v1 is zero.

With an initial velocity of v0 (upwards), and an acceleration due to gravity of -g (downwards), we use the relation
v1=v0+at
substituting v1=0, v0=16.5 and g=-9.81, you would be able to solve for t (answer to part a)
The remaining parts can be readily answered knowing t.

To find the time required for the football to reach the highest point in its trajectory, we can use the fact that the vertical velocity component at this point is zero. We can use the kinematic equation:

vf = vi + at

where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time. In this case, we have vf = 0 m/s, vi = 13.5 m/s (upward), and a = -9.8 m/s^2 (acceleration due to gravity).

Applying the equation:

0 = 13.5 - 9.8t

Simplifying and solving for t:

9.8t = 13.5
t = 13.5 / 9.8
t ≈ 1.38 seconds

So, it takes approximately 1.38 seconds for the football to reach the highest point in its trajectory.

To find how high the football gets above the ground, we need to determine the vertical displacement at the highest point, which we can calculate using the kinematic equation:

Δy = viy * t + (1/2) * a * t^2

where Δy is the vertical displacement, viy is the initial vertical velocity component, a is the acceleration, and t is the time. In this case, viy = 13.5 m/s (upward), a = -9.8 m/s^2 (acceleration due to gravity), and t ≈ 1.38 seconds.

Applying the equation:

Δy = (13.5)(1.38) + (1/2)(-9.8)(1.38)^2

Simplifying:

Δy ≈ 9.19 meters

Therefore, the football reaches a height of approximately 9.19 meters above the ground.

To find how much time after it is thrown it takes for the football to return to its original height, we can use the symmetry of the projectile motion. The time taken to reach the highest point is equal to the time taken to return to the original height.

So, it takes approximately 1.38 seconds for the football to return to its original height.

To find how far the football has traveled horizontally from its original position, we can use the horizontal velocity component and the time taken for the football to return to its original height.

Distance traveled = horizontal velocity * time taken

In this case, the horizontal velocity is 16.5 m/s (constant throughout), and the time taken is approximately 1.38 seconds.

Distance traveled ≈ 16.5 m/s * 1.38 s

Distance traveled ≈ 22.77 meters

Therefore, the football travels approximately 22.77 meters horizontally from its original position.