Answer :
The final velocity becomes 31.48 m/s
Explanation:
Given:
Initial velocity, u = 33 m/s
Height, h = 5m
Final velocity, v = ?
According to Newton's law:
v² - u² = 2gh
where,
g is the acceleration due to gravity and
g = 9.8 m/s²
On substituting the values we get:
[tex]v^2 - (33)^2 = 2 X -9.8 X 5\\\\v^2 - 1089 = -98\\\\v^2 = 991\\\\v = 31.48 m/s[/tex]
Therefore, the final velocity becomes 31.48 m/s
The velocity of the football at the given height is 31.48m/s.
Given the data in the question;
- Initial velocity; [tex]u = 33m/s[/tex]
- Height or distance; [tex]s = 5m[/tex]
Final velocity; [tex]v =\ ?[/tex]
To determine the velocity of the ball at the given height, we use the third equation of motion:
[tex]v^2 = u^2 + 2as[/tex]
Where v is final velocity, u is initial velocity, s is distance or height and a is acceleration due to gravity( since the ball is under gravity, [tex]a = g = -9.8m/s^2[/tex] ) The football is going upward, Acceleration due to gravity acts in the opposite direction of the velocity, so it becomes NEGATIVE.
We substitute our given values into the equation
[tex]v^2 = (33m/s)^2 + [ 2\ *\ -9.8m/s^2\ *\ 5m ]\\\\v^2 = 1089m^2/s^2\ -\ 98m^2/s^2\\\\v^2 = 991m^2/s^2\\\\v = \sqrt{991m^2/s^2}\\\\v = 31.48m/s[/tex]
Therefore; the velocity of the football at the given height is 31.48m/s.
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