Question

(6%) Problem 14: A football player punts the ball from the ground at a 45.0° angle above the horizontal. Without an effect from the wind, the ball would travel 63 m before falling back to the ground. 50% Part (a) What is the initial speed of the ball in m/s? Vo = 24.85 ✓ Correct! D 5 0% Part (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.3 m/s. What total distance, in meters, travels the ball horizontally before returning to the ground? x= Grade Summary Deductions 18% Potential 82% sin cos t an() ( ) 7 8 9 HOME Submissions cotan asin() acos) E 4 5 6 Attempts remaining: 1 atan acotan 1 sinh) * (3% per attempt) 1 2 3 detailed view cosh tanh cotanh() END Degrees Radians VO BACKSPACE DEL CLEAR + Submit Hint V Feedback I give up! Feedback: 0% deduction per feedback. Hints: 3 for a 0% deduction. Hints remaining: 0 - What is different about the motion before and after the wind gust? -How does the gust of wind change the time of flight? -Split the motion into two distinct parts, with a different horizontal velocity for each part.

Answer 1

given that

angle = 45 degrees

horizontal distance x = 63 m

from the relation

a) horizontal distance x = vi^2*sin(2*theta)/g

x = vi^2*sin(2*45)/9.8

63 = vi^2/9.8

vi = sqrt(63*9.8) = 24.847 m/s

b) horizontal velocity vx = vi*cos(45) = 24.847*cos(45) = 17.569 m/s

after wind blow

horizontal velocity vx = 17.569-1.3 = 16.269 m/s

total time t = 63/17.569 = 3.59 sec

total distance = 16.269*3.59 = 58.34 m