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asambeis [7]
3 years ago
5

If a ball is thrown up at an initial speed of 40. m/s, how many seconds does it take to reach the top of its path?

Physics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

Time, t = 4.08 secs

Explanation:

<u>Given the following data;</u>

Initial velocity, U = 40m/s

To find the time, we would use the first equation of motion;

V = U + at

Where;

  • V is the final velocity.
  • U is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

<em>Making time, t the subject of formula, we have;</em>

t = \frac{V - U}{a}

We know that acceleration due to gravity, g is 9.8m/s².

a = g = - 9.8m/s² because the ball is thrown in the opposite direction.  

Also, the final velocity is equal to zero (0) because the ball reached its maximum height.

<em>Substituting into the equation, we have;</em>

t = \frac{0 - 40}{-9.8}

t = \frac{-40}{-9.8}

Time, t = 4.08 secs

<em>Therefore, it will take the ball 4.08 seconds to reach the top. </em>

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