Answer to Question #90937 in Mechanics | Relativity for Manav

Question

A student throws balls outside the window at all possible directions with speed u. The pebbles hit the ground with an angle theta with the horizontal. Deduce the expression for the height of projection above ground level. According due to gravity is g

Answer 1

Assume that our student throws the balls with initial velocity $u$ at an angle $\alpha$ above the horizontal. Then if the α is positive, the ball s will be thrown upward or downward with negative $\alpha$. Also, assume that the window from where the student throws is $h_w$ meters above the ground. Begin deducing.

1) The height that the ball reaches when it moves upward will be defined by the vertical component of $u$ only:

$$h=\frac{(u\text{sin}\theta)^2 }{2g}=\frac{u^2\text{sin}^2\theta}{2g}.$$

Hence the total height will be

$$H=h_w+h$$

2) The vertical velocity of the ball thrown at some angle above or below the horizontal at any moment will be

$$v_v=u\space\text{sin}\alpha-gt.$$

We see that according to this expression with negative α the speed gradually increases or decreases and then increases if we begin counting time and measuring speed with positive α.

3) Meanwhile the horizontal component of the speed is constant if we neglect the air resistance:

$$v_h=u\space\text{cos}\alpha.$$

4) When the ball hits the ground, the angle theta can be defined as

$$\theta=\text{arctan}\frac{v_v}{v_h}=\text{arctan}\frac{u\space\text{sin}\alpha-gt}{u\space\text{cos}\alpha},$$

or in other words (expressing time in terms of other terms from the previous expression):

$$t=\frac{u(\text{sin}\alpha-\text{cos}\alpha\cdot\text{tan}\theta)}{g}.$$

5) The height at any moment:

$$H=h_w+h=h_w+\frac{1}{2}gt^2.$$

Substitute our $t$ and get the expression for the height above the ground:

$$H=h_w+\frac{u^2(\text{sin}\alpha-\text{cos}\alpha\cdot\text{tan}\theta)^2}{2g}$$