A 1.0kg football is given an initial velocity at ground level of 20.0m/s (37 degrees above horizontal). it gets blocked just after reaching a height of 3.0m
(a) what is the velocity of the football when it first reaches a height of 3.0m above the level ground?
(b) what horizontal distance has the ball travelled when it first reaches a height of 3.0m above ground?
To solve this problem, we need to use the equations of motion for projectile motion. We can break up the initial velocity into its horizontal and vertical components:
Vx = 20.0 cos(37) = 16.0 m/s
Vy = 20.0 sin(37) = 12.0 m/s
(a) To find the velocity of the football when it first reaches a height of 3.0m, we can use the equation for vertical displacement:
Δy = Voy*t + 0.5*a*t^2
where Δy is the vertical displacement (3.0m), Voy is the initial vertical velocity (12.0m/s), a is the acceleration due to gravity (-9.81m/s^2), and t is the time it takes for the ball to reach that height.
Rearranging the equation, we get:
t = (Vf - Voy)/a
where Vf is the final vertical velocity when the ball reaches 3.0m. At this point, the vertical velocity is zero, since the ball has reached the highest point of its trajectory. So we have:
0 = 12.0 - 9.81*t
t = 1.225s
Now we can use the equation for horizontal displacement:
Δx = Vx*t
where Vx is the initial horizontal velocity (16.0m/s) and t is the time it takes for the ball to reach 3.0m.
Δx = 16.0*1.225 = 19.6m
Therefore, the velocity of the football when it first reaches a height of 3.0m above the ground is:
Vf = Voy + a*t
= 12.0 - 9.81*1.225
= -1.11m/s
Note that the velocity is negative, indicating that the ball is moving downwards.
(b) The horizontal distance the ball has travelled when it first reaches a height of 3.0m above ground is:
Δx = Vx*t
= 16.0*1.225
= 19.6m
Therefore, the ball has travelled 19.6m horizontally when it first reaches a height of 3.0m above the ground.