In a cricket, a bowler throws a ball at the height 5m. A ball bounces on the pitch at 8m away. A wicket keeper catches this ball from 5m away at the height of 8m. How much distance the ball has covered after pitch. Please show the method.
Make a sketch and you should have two similar triangles each of sides 5 and 8.
Assumption: we will assume the ball travels in a straight line, and we will ignore the small actual curvature.
The sum of the two acute angles in each triangle = 90°
So the angle between the two paths of the ball is also 90°
Thus Pythagoras!
Dist^2 = (√89)^2 + (√89)^2 = 178
Dist = √178 = appr 13.34 m
To find the distance the ball has covered after the pitch, we need to calculate the horizontal distance the ball has traveled.
First, let's consider the horizontal motion. We can use the formula for horizontal distance (d) traveled by an object with initial velocity (u) and time (t):
d = u * t
In this scenario, the ball is thrown by the bowler at a height of 5m and bounces on the pitch 8m away. The wicket keeper then catches the ball at a height of 8m, which is 5m away horizontally.
So, the initial horizontal velocity (u) of the ball is the same as its final horizontal velocity. This means that the horizontal velocity remains constant throughout the motion.
We can assume that there is no air resistance, so the vertical motion (the height) does not have any effect on the horizontal motion.
Now, let's calculate the time it takes for the ball to travel 5m horizontally:
We can use the formula for horizontal distance with constant velocity:
d = u * t
5m = u * t
We know that the initial velocity (u) is the same as the final velocity, so we can substitute the value of the initial horizontal velocity (u):
5m = 8m/s * t
Now, solve for t by dividing both sides of the equation by 8m/s:
t = 5m / 8m/s
t = 0.625s
Therefore, it takes 0.625 seconds for the ball to travel 5 meters horizontally.
Now, let's find the total distance the ball has covered after the pitch. We can use the formula for total distance (s) traveled by an object with initial velocity (u), time (t), and constant acceleration (a):
s = u * t + (1/2) * a * t^2
In this scenario, there is no acceleration horizontally, so the horizontal distance traveled is simply:
s = u * t
s = 8m/s * 0.625s
s = 5m
Therefore, the ball has covered a total distance of 5 meters horizontally after the pitch.