Question

A rock is thrown with an initial speed of $30m/s$ at an angle of ${60}^o$ above the horizontal. It is thrown from a canyon floor and goes up into the air and then lands on the plateau, $20m$ above the canyon floor. How far does it travel horizontally? (mark the nearest integer)
($g=9.8m/s^2$)

• A. $46m$
• B. $60m$
• C. $27m$
• D. $30m$

Answer 1

The correct option is A $46m$

Maximum height of the projectile is given by $H=\frac{u^{2}Si{n}^2\theta }{2g}$

So $H=900\times \frac{3}{4}\times \frac{1}{2g}$.

$H=34.43meters$

Ball will first go to top then return to height 20 m(from the ground) or 14.43 m from top.

So total time will be $\frac{1}{2}T$ + time taken for the downward motion from the highest point.

$acceleration$ = $9.8\frac{m}{s^2}$

$displacement=14.43m$

$\frac{uSin60}{g}+\sqrt{\frac{2\times 14.43}{g}}$ $=>3.041$

So time taken will be 3.04. secs

Horizontal distance covered will be $uCos60\times$ time of flight

D=45.6m