A projectile is shot from the ground at a velocity of 15 m/s at an angle of pi/6. How long will it take for the projectile to land?

2 Answers
Dec 17, 2017

The time is =1.53s

Explanation:

Resolving in the vertical direction uarr^+

The initial velocity is u_0=15sin(1/6pi)ms^-1

The time to reach the greatest height is =ts

Applying the equation of motion

v=u+at=u- g t

The time is t=(v-u)/(-g)=(0-15sin(1/6pi))/(-9.8)=0.765s

The time taken to land is twice the time to reach the greatest height

T=2t=2*0.765=1.53s

Dec 17, 2017

1.531 s

Explanation:

Here you actually asked about the wandering period of the projectile.

The initial velocity,v_0=15 ms^(-1)

The throwing angle of the projectile,θ=pi/6=30@
Suppose,
The wandering period of the projectile, color(red)(T)=?

we know that,
T=color(blue)((2v_0sintheta)/g)=(2xx15ms^(-1)xxsin30@)/(9.8 ms^(-2))=1.531 scolor(red)((Ans.)