Question

If an object is projected at angle of Φ to the horizontal.it's maximum height is 8m.what is the time taken to reach the maximum height?.(g=10m\s)

Answer 1

The time is `=1.26s`

Explanation:

Resolving in the vertical direction `uarr^+`

The initial velocity is `u_0=(u)sin(phi)ms^-1`

The acceleration due to gravity is `a=-g=-10ms^-2`

Applying the equation of motion

`v^2=u^2+2as`

At the greatest height, `v=0ms^-1`

The greatest height is `h_y=s=0-((u)sin(phi))^2/(-2xx10)`

`h_y=(usin(phi))^2/(20)=8m`

Therefore,

`usinphi=sqrt(160)`

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-(u)sin(phi))/(-10)=sqrt(160)/10=12.65/10=1.26s`