- #1
bob1182006
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- 1
Homework Statement
A football player punts the football so that it will have a "hang time" (time of flight) of 4.50s and land 50 yd (=45.7m) away.
If the ball leaves the player's foot 5.0 ft (=1.52m) above the ground, what is it's initial velocity (magnitude and direction)?
Homework Equations
[tex]x=v_{0x}t[/tex]
[tex]v_x=v_{0x}[/tex]
[tex]v_y=gt[/tex]
[tex]v_0=v[/tex]
[tex]\theta_0=\theta+\pi[/tex]
[tex]v=\sqrt{v_x^2+v_y^2}[/tex]
[tex]\theta=arctan\frac{v_y}{v_x}[/tex]
The Attempt at a Solution
plugging into solve for x and y component of velocity I get:
[tex]v_x=v_{0x}=\frac{x}{t}=\frac{45.7m}{4.50s}=10.16 m/s[/tex]
[tex]v_y=gt=-9.81m/s^2 * 4.50s=-44.15m/s[/tex]
[tex]v=\sqrt{v_x^2+v_y^2}=\sqrt{(10.16)^2+(-44.15)^2}=\sqrt{2052}=45m/s[/tex]
[tex]\theta=arctan\frac{-44.15}{10.16}=-76.5[/tex] degrees
initial velocity =45 m/s
angle is 76.5 degree's.
but my book says this is wrong >< degrees should be ~65, velocity about 35 m/s
Am I using the wrong equations?