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a soccer ball is kicked and left the ground at angle 45° above the horizantal, moving at 25 m/s.
CALCULATE FOR THE:
a. hang time of the ball
b. initial horizontal velocity
c. initial vertical velocity
d. maximum height
e. the range travelled by ball

help me please :((​

Respuesta :

Trancescient

Answer:

a) 3.6 sec

b) 17.7 m/s

c) 17.7 m/s

d) 63.6 m

Explanation:

y = (25sin45)t - 4.9t^2 =17.7t - 4.9t^2

a) to find hang time set y = 0

4.9t^2 = 17.7t or t = 3.6 sec

b) vx = 25cos45 = 17.7 m/s

c) vy = 25sin45 = 17.7 m/s

d) To find the range R, use the results in (a) and (b) into the equation R = vxt:

R = (17.7 m/s)(3.6 sec) = 63.6 m

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