A hard ball dropped from a height of 1 m in earth’s gravitational field bounces to a height of 95 cm. What
will be the total distance traversed by the ball?
4 answers
What is 1m+.95m? This is a question in high school physics?
And horizontally the distance traversed is 0 m.
Or is the question trying to get you to add up the total distance travelled until it come to a standstil?
before bounce 2
So 1 m + 2x0.95
before bounce 3
So 1 m + 2x0.95 + 2x(0.95x0.95)
before bounce 4
So 1 m + 2x0.95 + 2x(0.95x0.95)+2x(0.95x0.95x0.95)
and so on to bounce n
1 m + 2x0.95 + 2x(0.95)^2+2x(0.95)^3 +...2x(0.95)^n-1
which I make tends towards 39 m, but check the maths.
Or is the question trying to get you to add up the total distance travelled until it come to a standstil?
before bounce 2
So 1 m + 2x0.95
before bounce 3
So 1 m + 2x0.95 + 2x(0.95x0.95)
before bounce 4
So 1 m + 2x0.95 + 2x(0.95x0.95)+2x(0.95x0.95x0.95)
and so on to bounce n
1 m + 2x0.95 + 2x(0.95)^2+2x(0.95)^3 +...2x(0.95)^n-1
which I make tends towards 39 m, but check the maths.
1m+(2*0.95m)+(2*0.9m)+(2*0.8m)+...+(2*.05m)+0
2m (2000cm)
2m (2000cm)
So the series is,
1 + 2(0.95) + 2(0.95)^2 + 2(0.95)^3 +...
all in meter. Which can be written as,
1 + 2*(Summation over n={1,infinity}(0.95^n)) = 1 + 2*(1/(1-0.95)-1)
Which comes to 39m
1 + 2(0.95) + 2(0.95)^2 + 2(0.95)^3 +...
all in meter. Which can be written as,
1 + 2*(Summation over n={1,infinity}(0.95^n)) = 1 + 2*(1/(1-0.95)-1)
Which comes to 39m
3 answers