Question

A projectile is launched ${60.0}^o$ above the horizontal with an initial speed of $15.0m/s$.
When the projectile returns to its original height some distance away, what will be the particle's speed?
(Take air resistance to be negligible and gravity to be the only source of acceleration)

• A. $7.50m/s$
• B. $13.0m/s$
• C. $15.0m/s$
• D. $22.0m/s$
• E. None of the above

Answer 1

The correct option is B $13.0m/s$

Given - $\theta ={60}^o,u=15.0m/s$ ,

therefore , initial vertical velocity $u_v=u\sin \theta$ ,

and initial horizontal velocity $u_h=u\cos \theta$ ,

or $u_h=15\times \sin 60=15/0.8660$ ,

or $u_h=13.0m/s$

at original height(maximum height) the vertical velocity of projectile becomes zero whereas horizontal velocity remains constant throughout the journey ,therefore total speed at original height will be due to only horizontal velocity that is $7.50m/s$ .