Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `15 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

The distance is `=11.5m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=15*sin(1/4pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=15sin(1/4pi)-g*t`

`t=15/g*sin(1/4pi)`

`=1.08s`

Resolving in the horizontal direction `rarr^+`

To find the horizontal distance, we apply the equation of motion

`s=u_x*t`

`=15cos(1/4pi)*1.08`

`=11.5m`

Answer 2

Here in our case
`theta=pi/4`
`u=15`m/s
`:.u_(y_1)=15sin(pi/4)=15/sqrt2`
`u_x=15cos (pi/4)=15/sqrt2`
When the projectile will reach maximum height
Velocity in `y` direction `u_(y_2)` `=0` we
`:.` Applying equating of motion along `y` direction we get
`(u_(y_2))^2 = (u_(y_1))^2 +2 (-g)h`
From this equation we get `h=5.74m`
Now time required to reach the highest point
`t=u_(y_2)/g`
`:.` distance moved in `x` ditection in this time
`d=u_(x)` `xxu_(y_2)/g`
`:.d=`11.5`m`
`:.`Distance from starting point of the projectile till it reaches the maximum height is given by

`l=sqrt (d^2+h^2)`
`:.l=12.8m`