Asked by musa
Science
A batsman hits a cricket ball from ground level. It is caught 1.8 seconds later by the third man standing 43.4 m from the bat. The ball leaves the bat at a speed of 26m/s. Calculate how high above the ground it is caught
A batsman hits a cricket ball from ground level. It is caught 1.8 seconds later by the third man standing 43.4 m from the bat. The ball leaves the bat at a speed of 26m/s. Calculate how high above the ground it is caught
Answers
Answered by
Ms. Sue
Your School SUBJECT is science.
Answered by
Steve
the x-velocity is 43.4/1.8 = 24.111 m/s
so, the ball is hit at an angle θ such that cosθ = 24.111/26, so θ = 21.97°
So, the y-velocity is 26sin21.97° = 9.73 m/s.
The height of the ball is
y = 9.73t - 4.9t^2
So, when caught after 1.8 seconds,
y(1.8) = 1.638
so, the ball is hit at an angle θ such that cosθ = 24.111/26, so θ = 21.97°
So, the y-velocity is 26sin21.97° = 9.73 m/s.
The height of the ball is
y = 9.73t - 4.9t^2
So, when caught after 1.8 seconds,
y(1.8) = 1.638
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.