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Komok [63]
3 years ago
14

A softball player moving 3.89 m/s

Physics
1 answer:
Ahat [919]3 years ago
3 0

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

Lets take final speed of the softball after covering 4.8 m = V

We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

Therefore slide time will be 0.119 s

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ale4655 [162]

Answer:

Explanation:

Given

Launch angle =u

Initial Speed is v_0

Horizontal acceleration is a_x=a

At maximum height velocity is zero therefore

v_f=v_i-gt

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\tan 2u=\frac{g}{a}

u=\frac{1}{2}tan ^{-1}\frac{g}{a}

(b)If a =10% g

a=0.1g

thus u=\frac{1}{2}tan^{-1}\frac{g}{0.1g}

u=42.14^{\circ}

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