SOLUTION: A batsman hits a cricket ball and the ball is caught 1,8 seconds later by a man standing at 43,4m from the bat. The velocity of the ball when it leaves the bat is 26m/s. Calculate
Algebra ->
Test
-> SOLUTION: A batsman hits a cricket ball and the ball is caught 1,8 seconds later by a man standing at 43,4m from the bat. The velocity of the ball when it leaves the bat is 26m/s. Calculate
Log On
Question 1038244: A batsman hits a cricket ball and the ball is caught 1,8 seconds later by a man standing at 43,4m from the bat. The velocity of the ball when it leaves the bat is 26m/s. Calculate how high above the ground will the ball go before being caught. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! A batsman hits a cricket ball and the ball is caught 1,8 seconds later by a man standing at 43,4m from the bat. The velocity of the ball when it leaves the bat is 26m/s. Calculate how high above the ground will the ball go before being caught.
------------
Horizontal speed Vh = 43.4/1.8 =~ 24.1111 m/sec
Vertical speed Vv = sqrt(26^2 - Vh^2) =~ 9.729 m/sec when hit
---
Using 5 m/sec/sec for gravity (you didn't spec a value):
h(t) = -5t^2 + 9.729t
The ball ascends for 0.9 sec (1/2 the flight time):
h(0.9) = 4.706 meters
