A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4m/s), (a) how far in advance of the recipients ( horizontal distance) must the goods be dropped. (b) Suppose, instead that the plane release the supplies a horizontal distance of 425m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climber's position? (c) with what speed do the supplies land in the latter case

Explanation:

For vertical displacement of food-supply

if time taken to reach the ground be t

235 = 1/2 g t²  , initial velocity downwards is zero.

t = 6.92 s

Horizontal displacement of food during this period

= 6.92 x 69.4 m

= 480.25 m .

b )

time required to cover horizontal distance of 425 m

= 425 / 69.4 = 6.124 s  

This time period will be the time of vertical fall of 235 m . Let initial vertical velocity required be u

h = ut + 1/2 gt²

235 = u x 6.124 + .5 x 9.8 x 6.124²

235 = u x 6.124 + 183.76

u = 8.36 m /s

c )

v = u + gt

= 8.36 + 9.8 x 6.124

= 68.37 m /s

This will be vertical component of velocity .

horizontal velocity = 69.4 m /s

resultant velocity = √ ( 68.37² + 69.4²)

= √(4674.45 +4816.36)

= 97.42 m /s

a person with a mass of 106kg has a weight of 1060n

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂