Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.

A flare fired upward at 49 m/s will reach a height of

Sagot :

Answer:

122.5 meters

Explanation:

We have the third equation of motion:

[tex]\boxed{\mathsf{v^2=u^2+2aS}}[/tex]

  • v stands for the final velocity
  • u, initial velocity
  • a is the acceleration of the particle in motion
  • S, the distance covered by the same particle.

Here, the particle is a flare.

It's initial velocity is 49 m/ s and the final will be zero, that's 'cause the particle comes to a standstill upon reaching its maximum, in case of motion under gravity.

a will be the acceleration due to gravity = 9.8 m/ s²

So, here are the terms we've got for the third eqn. of motion:

  • v = 0 m/ s
  • u = 49m/ s
  • a = - 9.8 m/ s (the force of gravity acts downwards, and that's the direction we say is negative)

Now, the term we've got to find out is the height reached by the flare, i. e., S.

=> v² = u² + 2aS

=> (0)² = (49)² + 2×(-9.8)×S

=> 0 = 49×49 - 2×9.8×S

=> 2×9.8×S = 49×49

[tex]\implies \mathsf{S=\frac{49\times 49}{2\times9.8} }[/tex]

=> S = [tex]\frac{245}{2}[/tex]

=> S = 122.5 meters

And, S is the maximum height the flare covers, so the answer is 122. 5 meters