Question #c6d6a

Your projectile is fired with an initial velocity of 22 m/s and at an angle of `33^@` with the horizontal.

`v_0 = "22 m/s"`

`theta=33^@`

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This means that the motion of the projectile has a vertical component and a horizontal component.

Vertically, the projectile will be acted upon by the gravitational acceleration, `g`. This means that you can write

`d_y = v_(0y) * t - 1/2 * g * t^2`, where

`d_y` - the total vertical displacement;
`v_(0y)` - the vertical component of `v_0`;
`t` - the total time of flight.

In your case `d_y` will be equal to zero because the projectile starts from the ground level and ends up on the ground level. The vertical component of `v_0` will be equal to

`v_(0y) = v_0 * sintheta`

The equation becomes

`0 = v_0 * (sintheta) * t - 1/2 * g * t^2`

`-1/2 * 9.8t^2 + 22 sin(33^@)t = 0`

`-4.9t^2 + 11.98t = 0 <=> t(-4.9t + 11.98) = 0`

The only acceptable solution is

`t = (-11.98)/-4.9 = color(green)("2.44 s")`

At the maximum vertical height, the vertical component of the velocity is zero. This means that you can write

`0 = v_(0y)^2 - 2 * g * h`, where

`h` - the maximum height the projectile reaches.

Solving this equation for `h` will get you

`h = (v_0^2sin(33^@)^2)/g = color(green)("7.32 m")`

Horizontally, the projectile is not acted upon by any forces. This means that you can write

`d_x = v_(0x) * t`, where

`d_x` - the range of the projectile;
`v_(ox)` - the horizontal component of `v_0`.

Use the values you have to solve for `d_x`

`d_x = v_0 * costheta * t = 22"m"/cancel("s") * cos(33^@) * 2.44cancel("s")`

`d_x = color(green)("45.0 m")`

it is defined as the time taken by any object to travel in any medium it is calculated by

t=2vsintheta/a
maximum vertical height is the peak of that projectile motion the object is travelling
horizontal range it is the range of horizontal motion which is done by object