A person in bus throws a ball straight up with a speed of 12 m/s. If the bus is moving at 25 m/s, what is the velocity of the ball to an observer on the ground? ( Ans. VR = 28 m/s, θR = 26° )


Given:

Initial vertical velocity of the ball = Vy = 12 m/s
The initial horizontal velocity of the ball is the same as the bus = Vx  = 25 m/s


To Find:

The resultant velocity of the ball = VR

Solution:

Since velocities Vx and Vy are perpendicular vectors to each other. So, to find the resultant velocity we will add these two velocity components together by making a right-angle triangle. By using the Pythagorean Theorem the magnitude of the resultant vector VR  will be:

VR  = `sqrt {Vx^2 + Vy^2}`

by putting values


VR = `sqrt {(25 m s⁻¹)^2 + (12 m s⁻¹)^2}`

VR = `sqrt {625 m² s⁻² + 144 m² s⁻²}`

VR = `sqrt {769 m² s⁻²}`

VR  = 27.7 m s⁻¹

VR  = 28 m/s ------------Ans. 1

Thus the magnitude of the resultant velocity is 28 m/s.

Now to determine the direction (angle) of the resultant velocity with horizontal component Vx, we are again using the perpendicular velocities components Vx and Vy. We know that;

tan θ = ` \frac {Vy}{Vx}`

tan θ = ` \frac {12 ms⁻¹}{25 m⁻¹}`

tan θ = 0.48

θ = tan⁻¹ 0.48

θ = 25.6°

θ = 26°  ----------------Ans. 2 

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