A football is kicked at ground level with a speed of 17.7 m/s at an angle of 39.8 ∘ to the horizontal.
How much later does it hit the ground?
Answer=> t=_____ s
Express your answer using three significant figures and include the appropriate units.
Vo = 17.7 m/s[39.8o].
Yo = 17.7*sin(39.8) = 11.33 m/s. = Vert. component of Vo.
Y = Yo + g*Tr = 0.
11.33 - 9.8Tr = 0, Tr = 1.16 s. = Rise time.
T = Tr+Tf. Tf = Tr.
T = Tr+Tr = 1.16 + 1.16 = 2.32 s.
To find the time it takes for the football to hit the ground, we can use the equations of motion in the vertical direction.
First, let's break down the initial velocity into its vertical and horizontal components:
Vertical component: V_y = V_initial * sin(theta)
Horizontal component: V_x = V_initial * cos(theta)
Given:
V_initial = 17.7 m/s
theta = 39.8 degrees
Using these values, we can calculate the vertical component of the initial velocity:
V_y = 17.7 m/s * sin(39.8 degrees)
V_y = 11.263 m/s (rounded to three significant figures)
Now, we can use the equation for vertical displacement to determine the time it takes for the ball to hit the ground:
Vertical displacement = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the ball is kicked at ground level, the initial vertical displacement is zero. Thus, the equation simplifies to:
0 = (1/2) * 9.8 m/s^2 * t^2
Rearranging the equation, we get:
9.8 m/s^2 * t^2 = 0
Since the product of any number and zero is zero, this equation is satisfied when t = 0. Therefore, the ball hits the ground immediately as it is kicked.
Answer: t = 0 s