A football is kicked 66.6 meters. If the ball is in the air 5.39 s, with what initial velocity was it kicked?
Tr = Tf = 5.39/2 = 2.695 s.
Tr = Rise time.
Tf = Fall time.
Y = Yo + gt.
Y = 0 + 9.8*2.695 = 26.4 m/s = Final velocity when falling = Initial velocity
(Yo)when rising. Y=Ver. componenttg of velocity.
Range = Xo*(Tr+Tf) = 66.6 m.
Xo*(5.39) = 66.6
Xo = 12.36 m/s. = Hor. component of
initial velocity.
tanA = Yo/Xo = 26.4/12.36 = 2.13592
A = 64.9o.
Vo=Xo/cosA=12.36/cos64.9=29.2 m/[email protected]
= Initial velocity.
To find the initial velocity at which the football was kicked, we can use the equation of motion for vertical motion:
s = ut + (1/2)gt^2
where:
s = displacement (66.6 m)
u = initial velocity (what we're trying to find)
t = time (5.39 s)
g = acceleration due to gravity (9.8 m/s^2)
Rearranging the equation, we have:
u = (s - (1/2)gt^2) / t
Substituting the given values, we can calculate the initial velocity:
u = (66.6 - (0.5 * 9.8 * 5.39^2)) / 5.39
u = (66.6 - (0.5 * 9.8 * 29.07)) / 5.39
u = (66.6 - 142.81) / 5.39
u = -76.21 / 5.39
u ā -14.14 m/s
Therefore, the initial velocity at which the football was kicked is approximately -14.14 m/s. The negative sign indicates that the ball was kicked in the downward direction.