Payton Manning throws a football with a velocity of 25.0 m/s to the horizontal. If the football strikes the floor 56.9 meters away horizontally from Payton, at what height above the ground did he let go of the football?
+118609 Let the co-ordinates of the start point be (0,0)
In the horizontal direction.
v=25m/s
How long does it take the ball to travel 56.9m
$$56.9m\times \frac{1s}{25m}=2.276sec$$
so t he ball is in the air for 2.276seconds
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initially in the vertical direction
t=0, v=0, a=-9.8
ongoing in the vertical direction
$$\\\ddot y=-9.8\\
\dot y=-9.8t+c_1\\
$Initial velocity = 0 so $c_1=0\\
\dot y=-9.8t\\
y=\frac{-9.8t^2}{2}+c_2\\
y=-4.9t^2+c_2\\
$When t=0, y=0 so $c_2=0\\
y=-4.9t^2\\$$
now find y when t=2.276
$${\mathtt{\,-\,}}\left({\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{2.276}}}^{{\mathtt{2}}}\right) = -{\mathtt{25.382\: \!862\: \!4}}$$
So initial the ball is 25.38metres above the ground (2 dec places)
+118609 Let the co-ordinates of the start point be (0,0)
In the horizontal direction.
v=25m/s
How long does it take the ball to travel 56.9m
$$56.9m\times \frac{1s}{25m}=2.276sec$$
so t he ball is in the air for 2.276seconds
--------------------------------------
initially in the vertical direction
t=0, v=0, a=-9.8
ongoing in the vertical direction
$$\\\ddot y=-9.8\\
\dot y=-9.8t+c_1\\
$Initial velocity = 0 so $c_1=0\\
\dot y=-9.8t\\
y=\frac{-9.8t^2}{2}+c_2\\
y=-4.9t^2+c_2\\
$When t=0, y=0 so $c_2=0\\
y=-4.9t^2\\$$
now find y when t=2.276
$${\mathtt{\,-\,}}\left({\mathtt{4.9}}{\mathtt{\,\times\,}}{{\mathtt{2.276}}}^{{\mathtt{2}}}\right) = -{\mathtt{25.382\: \!862\: \!4}}$$
So initial the ball is 25.38metres above the ground (2 dec places)
