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

Answer:

$\displaystyle y_m=3.65m$

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

$\displaystyle v_x=v_o\ cos\theta$

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

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

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

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

$\displaystyle t=\frac{v_o\ sin\theta }{g}$

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

$\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2$

Operating and simplifying

$\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}$

We have

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

The maximum height is

$\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}$

$\displaystyle y_m=3.65m$