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emmasim [6.3K]
3 years ago
10

Lacie kicks a football from ground level at a velocity of 13.9 m/s and at an angle of 25.0° to the ground. How long will the bal

l be in the air before it lands?
Round your answer to the nearest tenth.
Physics
1 answer:
alukav5142 [94]3 years ago
5 0
Just consider the vertical component
d_y=v_{0,y}t+\frac{1}{2}at^2
d_y=(13.9\sin 25\°)t+\frac{1}{2}(-9.8~m/s^2)t^2 \\  \\ d_y=5.87t-4.9t^2

Set d_y equal to 0 to find the times the ball is on the ground
0=5.87t-4.9t^2 \\  \\ 0=t(5.87-4.9t) \\  \\ t=0 \\  \\ t=1.2~s

The ball was in the air for 1.2 seconds
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The following mathematical operations are and are not allowed on two quantities with different unit dimensions are C) Allowed : Multiplication, Division , Not Allowed : Addition, Subtraction, Equality

To answer the question, we have to know what mathematical operations are.

<h3>What are Mathematical operations?</h3>

Mathematical operations are operations which are performed to change the value of a variable. The arithmetic mathematical operations we have are

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So, the mathematical operations that are allowed are multiplication and division while the mathematical operations that are not allowed are addition, subtraction and equality.

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Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

Initial vertical component of velocity:

\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}.

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing y_{1} is the same as the altitude y_{0} at which this projectile was launched: y_{0} = y_{1}.

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is 20.6\; {\rm m\cdot s^{-1}} (upwards,) the vertical velocity right before landing would be (-20.6\; {\rm m\cdot s^{-1}}) (downwards.) The change in vertical velocity is:

\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

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