A person standing on the top of a cliff 171 ft high has to throuw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend wilth a speed o 15.0 ft/s, how short will the packet fall?

Answer

B, C

Explanation

Given, `tan theta` = 171/228
`rarr theta = tan^-1 (171/228)`
The motion of projectile (i.e the packet) is from A. Take reference axis of a.
`:. Theta = 37^0` as u is below x-axis`
u= 15 t. s, g= 32.2 ft/s^2, y=- 171 ft

Form `y= tan theta x- (x^2 g sec^2 theta)/(2u^2)`
`-171= - (0.7536) x- (x^2 g(1.568)/(2(225))`
`0.1125x^2-0.7536X- 171=0`
`rarr x= 35.78 ft (can be calculated)`
Horizontal range covered by the packet is 35.78 ft.
So, the packet will fall
228-35.78
182 ft short of his friend.