Question

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?

Answer 1

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`

Answer 2

`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 s` `color(red)((Ans.)`