Physics, asked by mahekmughal1301, 8 months ago

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?

Answers

Answered by princi58

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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)

Answered by bhuvna789456

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.

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