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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?

User Kiran K G
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1 Answer

2 votes

Answer:

(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.

User Robert Newton
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