A football is thrown with an initial upward velocity component of 15.0m/s and a horizontal velocity component of 18.0m/s .
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?
The first two questions are dependent on the vertical velocity
how high? at the top vfinal is zero.
find the time it takes to get to that point where vf is zero.
vf=vi+gt
then, using that time,
h=vi*t+1/2 gt^2 where g is -9.8
how far? use the horizontal velocity
distance=horizvel*timeinair.
t=1.53
complete pass time = 3.06
h= 30.4m
d(horizontal total)= 55 m/s
To solve these questions, we need to use the kinematic equations of motion. Here's how to find the answers step by step:
1. How high does the football get above the ground?
We can use the vertical motion equation:
h = (v0y^2) / (2 * g),
where h is the maximum height, v0y is the initial upward velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
v0y = 15.0 m/s.
Plugging in the values:
h = (15.0^2) / (2 * 9.8).
h = 225 / 19.6.
h ≈ 11.48 m.
Therefore, the football reaches a maximum height of approximately 11.48 meters above the ground.
2. How much time after it is thrown does it take to return to its original height?
Since the vertical motion of the football is symmetrical (assuming no air resistance), it will take the same amount of time to reach its maximum height as it takes to return to its original height.
To find this time, we can use the equation:
t = v0y / g,
where t is the time, v0y is the initial upward velocity, and g is the acceleration due to gravity.
Given:
v0y = 15.0 m/s.
Plugging in the values:
t = 15.0 / 9.8.
t ≈ 1.53 seconds.
So, it takes approximately 1.53 seconds for the football to return to its original height.
3. How far has the football traveled horizontally from its original position?
We can use the horizontal motion equation:
d = v0x * t,
where d is the horizontal displacement, v0x is the initial horizontal velocity, and t is the time.
Given:
v0x = 18.0 m/s,
t ≈ 1.53 seconds (from previous calculation).
Plugging in the values:
d = 18.0 * 1.53.
d ≈ 27.54 meters.
Therefore, the football has traveled approximately 27.54 meters horizontally from its original position.