Question

A person standing on the top of a cliff $171ft$ high has to throw a packet to his friend standing on the ground $228ft$ horizontally away. If he throws the packet directly aiming at the friend with a speed of $15\cdot 0ft/s$, how short will the packet fall?

Answer 1

The straight line distance from A to B is

$\sqrt{{171}^2+{228}^2}$ = 285

feet.

Packet is thrown along the direction AB with speed 15 ft/s.

If we resolve this speed into vertical and horizontal components,

we get vertical component as 9 ft/s

( 15 sinθ = $\frac{15\times 171}{285}$ = 9 ft/s)

and horizontal component as 12 ft/s

$(15cos\theta =\frac{15\times 228}{285}$ = 12 ft/s )

The time to reach the ground for the packet is decided by gravity.

Hence we will find the time taken to reach the ground using vertical component.

we need to use the equation of motion, " S = $ut+\left(\frac{1}{2}\right)g{t}^2$ "

171 = $9\times t+\left(\frac{1}{2}\right)\times 9.8\times 3.281\times t^2$

$=9\times t+16.08\times t^2$ ...................(1)

(acceleration due to gravity is $ft/s^2$ . 1 m = 3.281 ft)

solving the quadratic equation (1) for t, we get t = 3 s.

Hence horizontal distance travelled by the packet = $12\times$3 = 36 ft.

Packet will fall at a distance 36 ft from O,

short of 192 ft to B ( 228-36 = 192 ft )