Agolf ball was thrown with a speed of 25 m/sec at an angle of 65° above the horizontal. what is the horizontal distance traveled by the ball before it hits the ground?

The horizontal distance traveled by the ball before it hits the ground is 48.85 meters.

Explanation:

It is given that,

Speed of golf ball, v = 25 m/s

Angle above horizontal or angle of projection, θ = 65°

We need to find the distance travelled by the ball before it hots the ground or in other words we need to find the range. It is given by R.

R = 48.85 m

So, the distance travelled by the ball before it hots the ground is 48.85 meters. Hence, this is the required solution.

  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

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