Question

You are part of a searchand- rescue mission that has been called out to look for a lost explorer. You’ve found the missing explorer, but you're separated from him by a 200-m m -high cliff and a 30-m m -wide raging river. To save his life, you need to get a 5.8 kg k g package of emergency supplies across the river. Unfortunately, you can't throw the package hard enough to make it across. Fortunately, you happen to have a 0.90 kg k g rocket intended for launching flares. Improvising quickly, you attach a sharpened stick to the front of the rocket, so that it will impale itself into the package of supplies, then fire the rocket at ground level toward the supplies. (Figure 1) Figure1 of 1A figure shows an explorer stranded across a 30-meter-wide river at the bottom of a 200-meter-height cliff. A rescuer attempts to deliver a package to the explorer by placing the package at the edge of the cliff and shooting a rocket horizontally at it. A figure shows an explorer stranded across a 30-meter-wide river at the bottom of a 200-meter-height cliff. A rescuer attempts to deliver a package to the explorer by placing the package at the edge of the cliff and shooting a rocket horizontally at it. Part A What minimum speed must the rocket have just before impact in order to save the explorer’s life? Express your answer to two significant figures and include the appropriate units.

Answer 1

Part A.

Given that rocket collides with package horizontally, So Initial vertical velocity of combined package and rocket is 0 m/sec

Vertical distance traveled by package + rocket = -200 m (since in downward direction)

Using 2nd kinematic equation in vertical direction:

H = V0y*t + (1/2)*a*t^2

-200 = 0*t + (1/2)*(-9.81)*t^2

t = sqrt (2*200/9.81)

t = 6.386 sec = time for which package is in the air

We know that range in projectile motion is given by:

R = V0x*t

V0x = Initial horizontal velocity of package + rocket

R = Range of package + rocket = 30 m (since we need minimum speed)

So,

V0x = R/t = 30/6.386

V0x = 4.70 m/sec = x-velocity of package + rocket

Since Package and rocket collides completely inelastically, So

Using momentum conservation before and after collision

Pi = Pf

m1*U1 + m2*U2 = M*V0

m1 = mass of package = 5.8 kg

m2 = mass of rocket = 0.90 kg

M = mass of package + rocket = 5.8 kg + 0.90 kg = 6.7 kg

U1 = Initial velocity of package = 0 m/sec

U2 = Initial velocity of rocket = ?

So,

5.8*0 + 0.9*U2 = 6.7*4.70

U2 = 6.7*4.70/0.9 = 34.989 m/s

In two significant figures

U2 = 35 m/sec = minimum Initial velocity of rocket

Let me know if you've any query.