Question

A ball 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?

Answer 1

Step-by-step 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

`v_(y)=u\cos\theta`

Put the value into the formula

`v_(y)=19.7\cos45`

`v_(y)=13.93\ m/s`

The top of the trajectory is 13.93 m/s.

(b). We need to calculate the maximum height

Using formula of height

`h=((u\sin\theta)^2)/(2g)`

Put the value into the formula

`h=((19.7\sin45)^2)/(2*9.8)`

`h=9.90\ m`

The maximum height is 9.90 m.

(c). We need to calculate the range

Using formula of range

`R=(u^2\sin2\theta)/(g)`

Put the value into the formula

`R=(19.7^2*\sin2*45)/(9.8)`

`R=39.6\ m`

The range is 39.6 m.

(d). We need to calculate the time

Using formula of time

`t=(2u\sin\theta)/(g)`

Put the value in to the formula

`t=(2*19.7\sin45)/(9.8)`

`t=2.84\ sec`

The time is 2.84 sec.

Hence, This is the required solution.