An airplane flying upward at 39.6m/s and an angle of 30 degree relative to the horizontal releases a ball when it is 287m above the ground. Calculate the time it takes the ball to hit the ground? Calculate the maximum height of the ball? Calculate the horizontal distance the ball travels from the release point to the ground? Neglect any effects due to air resistance.

Assuming the speed of the plane is 39.6 m/s

vy = 39.6 * sin 30 = 19.8 vertical speed

vx = 39.6 * cos 30 = 34.3 m/s

S = vy t - 1/2 a t^2    to calclulate vertical height

-287 = 19.8 t - 4.9 t^2

4.9 t^2 -19.8 t - 287 = 0

I get t = 9.93 for solving the quadratic for time for ball to hit ground

Horizontal distance = 34.3 m/s * 9.93 s = 341 m

With a vertical speed of 19.9 m/s it takes 19.8 / 9.8 = 2.02 s to reach max height

You can again solve a quadratic or use 19.8 / 2 * 2.02 for time to reach vertical height         H = 287 + 20 = 307 m

As a check 307 = 1/2 g t^2 = 4.9 * (9/93 - 2/02)^2 = 307 m  max height

After you know how long the ball is in the air there are several ways to calculate the other values.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz