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labwork
1 month ago
10

At what height h above the ground does the projectile have a speed of 0.5v?

Physics
1 answer:
Yuliya22 [3.3K]1 month ago
6 0

Answer:

h=\dfrac{3v^2}{8g}

Explanation:

It is stated that

The speed of the projectile is 0.5 v. Let h denote the height above the ground. We will use the first equation of motion to calculate it.

v=u+at

v=u-gt

The initial speed of the projectile is v and it final speed is 0.5 v.

0.5v=v-gt

t=\dfrac{v}{2g}

g represents the acceleration due to gravity

Let h be the height above ground. By applying the second equation of motion, we have:

h=vt-\dfrac{1}{2}gt^2

h=v\dfrac{v}{2g}-\dfrac{1}{2}g(\dfrac{v}{2g})^2

h=\dfrac{3v^2}{8g}

Thus, the height of the projectile above the ground is \dfrac{3v^2}{8g}. Consequently, this is the solution we need.

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