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blondinia [14]
4 years ago
12

Can some answer 7 a and b please

Physics
1 answer:
pav-90 [236]4 years ago
4 0

Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

V_{f} =0.8 + 0.06 \times 15

V_{f} =0.8+ 0.9

V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

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Answer and Explanation:

As per the question:

When the stone is thrown from the cliff top and hits the ground below eventually:

R = v_{o}\sqrt{\frac{2H}{g}}

where

v_{o} = initial velocity

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Now,

(a) Displacement of the stone is given by the horizontal range:

R = v_{o}\sqrt{\frac{2H}{g}}

where

v_{o} = initial velocity

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v = \sqrt{v_{o}^{2} + 2gH}

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v = final velocity

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H = v_{o}T + \frac{1}{2}gT^{2}

H = 0.T + \frac{1}{2}gT^{2}

T = \sqrt{\frac{2H}{g}

3 0
3 years ago
Will give brainliest to right answer!
viva [34]
The correct answer is D.
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4 years ago
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Explanation:

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