Question

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

Answer 1

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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+(1/2)gt2 " 171 = 9×t+(1/2)×9.8×3.281×t2 = 9×t+16.08×t2  ...................(1) 

 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+(1/2)gt2 " 171 = 9×t+(1/2)×9.8×3.281×t2 = 9×t+16.08×t2  ...................(1) (acceleration due to gravity is ft/s2. 1 m = 3.281 ft)

Answer 2

The packet will fall  =228-35.8 = 192 ft  short of his friend.

Explanation:

Step 1:    

Given as

$\tan \theta=\frac{171}{228}$

$\begin{aligned}&\theta=\tan ^{-1} \frac{171}{228}\\&\theta=\tan ^{-1} 0.75\end{aligned}$

Θ = 36.8°  

The motion of projectile is from point A. Take reference axis at a.

θ=37° as u is below x axis.

Step 2:

he tosses the bundle straightforwardly focusing on the companion with a speed of 15.0 ft/s  

u= 15 ft/s

$g=32.2 f t / s^2$

y = -171 ft

$\text { From } y=\tan \theta-x^{2} \frac{g}{2 u^{2}}(\sec \theta)$

$\begin{array}{l}{-171=\left(\tan 37^{\circ}\right) x-x^{2} \times \frac{32.2}{2 \times 15^{2}} \times\left(\sec 37^{\circ}\right)} \\\\{-171=(0.753) x-x^{2} \times \frac{32.2}{2 \times 225} \times(1.252)}\end{array}$

$\begin{aligned}&-171=0.753 x-1.252 x^{2} \times \frac{32.2}{450}\\&-171=0.753 x-1.252 x^{2} \times 0.0715\end{aligned}$

$-171=0.753 x-0.089 x^{2}$

0.089~0.1

$0.1 x^{2}-0.753 x-171=0$

Step 3:

After solve the above equation  

we can get  that x=35.78 ft

Horizontal range covered by packet is 35.8 ft .

So the packet will fall  =228-35.8 = 192 ft  short of his friend.