A football is kicked across a level field. The ball spends 3.9 seconds in the air and lands a distance X = 49 meters from the point where it was kicked. The initial speed of the ball is V0. You should ignore air resistance. 1)What is the speed of the ball at the top of its trajectory?

Explanation:

At the top of the trajectory , the velocity of the projectile is equal to the horizontal component of the initial velocity .

Given the range of the projectile is 49 m and its time of flight is 3.9 s that means it covers the range in 3.9 s . Horizontal displacement is equal to range which is 49 m and horizontal displacement is due to horizontal component of initial velocity .

So horizontal component of initial velocity = horizontal displacement / time

= 49 / 3.9 = 12.56 m /s

Hence velocity at the top = 12.56 m /s .

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//