Question

A football player punts the ball at a 49° angle. Without an effect from the wind the ball would travel 62.0 meters horizontally.

16. +/2 points OSColPhys1 3.4.047 My Notes Ask Your Teacher A football player punts the ball at a 49° angle. Without an effect from the wind the ball would travel 62.0 meters horizontally. (a) What is the initial speed of the ball? m/s (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.90 m/s. What distance does the ball travel horizontally? Additional Materials Reading Submit Answer Save Progress Practice Another Version

Answer 1

1)

along horizantal

initial velocity vox = vo*costheta

ax = 0

from equaiton of motion

x = vox*T+ 0.5*ax*T^2

x = vo*costheta*T

T = x/(vo*costheta)......(1)

along vertical

voy = vo*sintheta

acceleration ay = -g = -9.8 m/s^2

displacement y = -30 m

from equation of motion

dy = voy*T + 0.5*ay*T^2

dy = (vo*sintheta*x)/(vo*costhetra) - (0.5*g*x^2)/(vo^2*(costheta)^2)

dy = x*tanthheta - ((0.5*g*x^2)/(vo^2*(costheta)^2))

after the ball reaches the ground dy = 0

x = 62 m

theta = 49

0 = (62*tan49) - ((0.5*9.8*62^2)/(v0^2*(cos49)^2))

v0 = 25.0 m/s

____

along vetical

for x = 62/2 = 31 m the ball reaches maximum height

H = (31*tan49) - ((0.5*10*31^2)/(25^2*(cos49)^2))

H = 18 m

at the maximum height

along horizantal

initial velocity vx = 1.9 m/s

acceleration ax = 0

along vertical

initial velocity = vy = 0

acceleration ay = -g m/s^2

H = 0.5*g*t^2

t = sqrt(2H/g)

along horizantal

X1 = vx*t = 1.9*sqrt((2*18)/10)

x1 = 3.6 m

total horizantal distance = 31 + 3.6 = 34.6 m