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?
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.
1m+(2*0.95m)+(2*0.9m)+(2*0.8m)+...+(2*.05m)+0
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
To determine the total distance traversed by the ball, we need to calculate the sum of the distances covered during both the upward and downward paths.
First, let's calculate the distance covered during the upward path:
The ball is dropped from a height of 1 m and reaches a height of 95 cm after bouncing.
Therefore, the distance covered in the upward path is the difference between the initial height (1 m) and the maximum height reached after bouncing (95 cm).
Converting the heights to meters, we have:
Initial height = 1 m
Maximum height reached = 95 cm = 0.95 m
Distance covered in upward path = Initial height - Maximum height reached
= 1 m - 0.95 m
= 0.05 m (or 5 cm)
Now let's calculate the distance covered during the downward path:
During the downward path, the ball covers the same distance as the upward path, plus the additional distance from the maximum height reached to the ground.
Distance covered in downward path = Distance covered in upward path + Additional distance to the ground
The additional distance to the ground is calculated as the difference between the initial height (1 m) and the maximum height reached (0.95 m).
Additional distance to the ground = Initial height - Maximum height reached
= 1 m - 0.95 m
= 0.05 m (or 5 cm)
Distance covered in downward path = Distance covered in upward path + Additional distance to the ground
= 0.05 m + 0.05 m
= 0.1 m (or 10 cm)
Finally, to find the total distance traversed by the ball, we add the distances covered during both the upward and downward paths:
Total Distance = Distance covered in upward path + Distance covered in downward path
= 0.05 m + 0.1 m
= 0.15 m (or 15 cm)
Therefore, the total distance traversed by the ball is 0.15 meters (or 15 centimeters).