Question

A ball is thrown with a speed of 20 m/s at an angle of 40o above the horizontal from the top of a 22-m tall building.

It hits a neighboring building 3.25 s later.

(a) How far away is the building, and (b) how fast was the ball going when it hit?

Answer 1

(a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.

Step-by-step explanation:

Given that,

Speed = 20 m/s

Angle = 40°

Height = 22 m

Time = 3.25 sec

(a). We need to calculate the distance

Using formula of distance

`d=u\cos\theta* t`

Put the value into the formula

`d=20\cos40*3.25`

`d=49.79\ m`

(b). We need to calculate the horizontal velocity

Using formula of velocity

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

Put the value into the formula

`v_(x)=20*\cos40`

`v_(x)=15.3\ m/s`

We need to calculate the vertical velocity

Using equation of motion

`v_(y)=u\sin\theta-gt`

Put the value into the formula

`v_(y)=20\sin40-9.8*3.25`

`v_(y)=-19\ m/s`

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

`v=\sqrt{v_(x)^2+v_(y)^2}`

`v=√((15.3)^2+(19)^2)`

`v=24.39\ m/s`

Hence, (a). The distance is 49.79 m.

(b). The speed of the ball is 24.39 m/s.