Question

If a bullet is fired from a point making an angle of 40° to the horizontal H. The initial velocity of bullet is 30m/s what will be it greatest highet reacted and the time to takes to reach maximum higher?

Answer 1

The bullet will go up due to its vertical component of velocity(`u sin theta`),and will move forward due to its horizontal component.

Given, `u=30 ms^-1`

So,if it reaches a maximum height of `h`,we can say

`0^2=(u sin 40)^2 -2gh`(using, `v^2=u^2 -2gh` here, at the highest point vertical component of velocity `v` becomes zero)

So, `h = (30 sin 40)^2/(2×g)=18.95m`

And,if it takes time `t` to reach that point,we can use, `0=u sin 40-g t` (using, `v=u-g t`)

So, `t = (u sin 40)/g=1.96 s`

Answer 2

`Deltax~~25.49m` and `t~~2.69s`

Explanation:

Because the bullet is shot at 40 degrees it means that there is a combination of velocity in the x and y direction. Because we are looking for height, we only need to deal with the velocity in the y direction. To separate the velocity of the bullet just in the y direction, we multiply the magnitude of the velocity by the sine of the angle:

`V_oy=30"m/s"*sin(40)`

=

`V_oy=19.284"m/s"`

We now bring out a kinematic equation. Assuming we are on earth and air resistance can be ignored, we can use the equation:

`Deltax=((V_y)^2-(V_oy)^2)/(2a_y)`

Where

`Deltax` is are height

`V_y` is the final velocity which we know will be zero at the greatest height due to the negative acceleration

`V_oy=19.284"m/s"`

`a_y` is are constant acceleration in the down direction on earth `(-9.8m/s^2)`

Plugging in our values for `Deltax` we get:

`Deltax=((0)^2-(19.284)^2)/(2(-9.8))`

`Deltax~~18.973m`

Now that we have `Deltax` we can us another kinematic equation that includes time.

`Deltax=V_oy*t+1/2*a*t^2`

Plugging in same values:

`18.973m=19.284"m/s"*t+(-4.9)*t^2`

=

`t~~1.963`