Aball is launched from the ground at an angle of 45° with an initial velocity of 19.7 m/s. (a) what is its velocity at the top of the trajectory? (remember that it has both and x and y components when it leaves the ground) (b) what is its maximum height? (c) what is its range? (d) how long was it in the air?

Explanation:

Given that,

Initial velocity = 19.7 m/s

Angle = 45°

(a). We need to calculate the velocity at the top of the trajectory

Using formula of velocity at the top of the trajectory

Put the value into the formula

The top of the trajectory is 13.93 m/s.

(b). We need to calculate the maximum height

Using formula of height

Put the value into the formula

The maximum height is 9.90 m.

(c). We need to calculate the range

Using formula of range

Put the value into the formula

The range is 39.6 m.

(d). We need to calculate the time

Using formula of time

Put the value in to the formula

The time is 2.84 sec.

Hence, This is the required solution.

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