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Brrunno [24]
3 years ago
9

You launch a baseball through the air at speed 37 m/sec, at angle 45 degrees above horizontal. Neglect air resistance. What shap

e is the trajectory?
Physics
1 answer:
Marizza181 [45]3 years ago
6 0

Answer:

Parabolic motion

Explanation:

It is given that,

Speed with which a baseball is launched through the ar, v = 37 m/s

Angle of projection, \theta=45^{\circ}

We know that when an object is projected with some initial speed having some angle of projection, the path followed by the object is called projectile motion. The object will act as a projectile here. As a result the shape of the trajectory will be parabolic.

The equation of parabolic path is given by :

y=tan\theta\ x-\dfrac{gx^2}{2u^2\ cos^2\theta}

It is similar to,

y=ax+bx^2 which is the equation of parabola

Hence, this is the required solution.                                                                                                    

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Answer:

The time taken for the race is 17.20 s.

Explanation:

It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.

Calculate the final speed of the sprinter.

The expression for the equation of the motion is as follows;

v^{2}-u^{2}= 2as

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

v^{2}-(0)^{2}= 2(1.44)(30)

v= 9.3 m^{-1}

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s=ut+\frac{(1)}{2}at^2

Put u= 0, s=30 m and a= 1.44 ms^{-2}.

30=(0)t+\frac{(1)}{2}(1.44)t^2

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T= t+t'

Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

T=6.45+\frac{s'}{v}

Put s= 30 m, v= 9.3 m^{-1} and s'= 100 m.

T=6.45+\frac{(100)}{9.3}

T= 17.20 s

Therefore, the time taken for the race is 17.20 s.

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