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A quarterback throws a football toward a re-ceiver with an initial speed of 17 m=s, at an
angle of 37 above the horizontal. At that
instant, the receiver is 17 m from the quarter-
back.
The acceleration of gravity is 9:8 m=s2 :
With what constant speed should the receiver run in order to catch the football at the
level at which it was thrown? Answer in units
of m=s.

2007-11-29 09:35:08 · 1 answers · asked by Jarod S 2 in Science & Mathematics Physics

1 answers

A typical projectile motion problem. Let the origin be the point of release of the football. +x is downfield; + is up. We assume that the catcher catches the ball at about the same level from which it was thrown.

Vertical motion:

vy = voy - gt

At the instant of the catch, vy = -voy (the ball is coming down at the same vertical speed with which it was thrown) so

2 voy / g = t

2 ( 17 m / s ) ( sin 37 degrees ) / 9.8 m / s^2 = t

Now we know the time. During that interval the ball travels a distance of

x - xo = vox t

vox is 17 m/s cos 37 degrees. Plug in the time and find the distance.

Now you know where the ball will be in t seconds and where the intended reciever is. The reciever must cover that difference in t seconds to make the catch.

2007-11-29 09:41:50 · answer #1 · answered by jgoulden 7 · 0 0

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