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Chapter 3: Q15. (page 67)

A projectile is launched at an upward angle of 30° to the horizontal with a speed of 30 m/s. How does the horizontal component of its velocity 1.0 s after the launch compare with its horizontal component of velocity 2.0 s after the launch, ignoring air resistance? Explain.

Short Answer

Expert verified

The required value of velocity is vx=25.98m/s.

Step by step solution

01

Step 1. Projectile motion

In order to determine the horizontal component of velocity in this problem, the air resistance is considered negligible.

Given data:

The horizontal speed of the projectile is vh=30m/s.

The angle made by the projectile is θ=30°.

The initial time after the launch is t1=1 s.

The final time after the launch is t2=2 s.

The velocity’s horizontal component remains constant in a projectile, motion which means that the component of the velocity of the projectile will be identical for both launch time

t1and t2.

02

Step 2. Calculation of the horizontal component of the velocity

The relation to calculate the horizontal velocity is given by

vx=vhcosθ.

On plugging the values in the above relation,

vx=30m/scos30°vx=25.98m/s

Thus, vx=25.98m/sis the horizontal component of the velocity.

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