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White raven [17]
3 years ago
9

A dart is thrown at a target that is supported by a wooden backstop. It strikes the backstop with an initial velocity of 350 m/s

[E] The dart comes to rest in 0.0050 s.
a. What is the acceleration of the dart?
b. How far does the dart penetrate into the backstop?​
Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
6 0

Answer:u=350m/sec

v=0m/sec

t=0.005sec

a=v-u/t

=0-350/0.005

= -350/0.005

= -350×1000/5

= -70×1000

= -70000 m/sec^2

Explanation: i think this will help you thankyou

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Answer:

t = 1.82

Explanation:

Given

u = 7.70m/s -- initial velocity

s = 30.2m --- height

Required

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Where

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So, we have:

30.2 = 7.70t + \frac{1}{2} * 9.8 * t^2

30.2 = 7.70t + 4.9 * t^2

Subtract 30.2 from both sides

30.2 -30.2  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9 * t^2 - 30.2

0  = 7.70t + 4.9t^2 - 30.2

7.70t + 4.9t^2 - 30.2  = 0

4.9t^2 + 7.70t - 30.2  = 0

Solve using quadratic formula:

t = \frac{-b\±\sqrt{b^2 - 4ac}}{2a}

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a = 4.9;\ b = 7.70;\ c = -30.2

t = \frac{-7.70\±\sqrt{7.70^2 - 4*4.9*-30.2}}{2*4.9}

t = \frac{-7.70\±\sqrt{651.21}}{9.8}

t = \frac{-7.70\±25.52}{9.8}

Split the expression

t = \frac{-7.70+25.52}{9.8} or t = \frac{-7.70-25.52}{9.8}

t = \frac{17.82}{9.8} or t = -\frac{33.22}{9.8}

Time can't be negative;  So, we have:

t = \frac{17.82}{9.8}

t = 1.82

Hence, the time to hit the ground is 1.82 seconds

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Answer:

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Explanation:

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