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.

Answer 1

Answer:

(a). The distance is  49.79 m.

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

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

$d=u\cos\theta\times t$

Put the value into the formula

$d=20\cos40\times3.25$

$d=49.79\ m$

(b). We need to calculate the horizontal velocity

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=20\times\cos40$

$v_{x}=15.3\ m/s$

We need to calculate the vertical velocity

Using equation of motion

$v_{y}=u\sin\theta-gt$

Put the value into the formula

$v_{y}=20\sin40-9.8\times3.25$

$v_{y}=-19\ m/s$

Negative sign shows the opposite direction.

We need to calculate the speed of ball

Using formula of speed

$v=\sqrt{v_{x}^2+v_{y}^2}$

$v=\sqrt{(15.3)^2+(19)^2}$

$v=24.39\ m/s$

Hence, (a). The distance is  49.79 m.

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