Question

An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the maximum height reached?​

Answer 1

`\displaystyle y_m=3.65m`

Step-by-step explanation:

Motion in The Plane

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of
`v_o` and
`\theta\\` as the initial speed and angle, then we have

`\displaystyle v_x=v_o\ cos\theta`

`\displaystyle v_y=v_o\ sin\theta-gt`

`\displaystyle x=v_o\ cos\theta\ t`

`\displaystyle y=v_o\ sin\theta\ t -(gt^2)/(2)`

If we want to know the maximum height reached by the object, we find the value of t when
`v_y` becomes zero, because the object stops going up and starts going down

`\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt`

Solving for t

`\displaystyle t=(v_o\ sin\theta )/(g)`

Then we replace that value into y, to find the maximum height

`\displaystyle y_m=v_o\ sin\theta \ (v_o\ sin\theta )/(g)-(g)/(2)\left ((v_o\ sin\theta )/(g)\right )^2`

Operating and simplifying

`\displaystyle y_m=(v_o^2\ sin^2\theta )/(2g)`

We have

`\displaystyle v_o=20\ m/s,\ \theta=25^o`

The maximum height is

`\displaystyle y_m=((20)^2(sin25^o)^2)/(2(9.8))=(71.44)/(19.6)`

`\displaystyle y_m=3.65m`