Question

A shell is fired at a angle at 60 degree to the horizontal direction with a velocity of 392 m/s. find
1- range
2 - height

this question is from porjectile motion​

Answer 1

I solved it in a piece of paper to be more clear

Answer 2

The range and height of the projectile are 13579.27 m and 5880 m respectively.

Explanation:

It is given that,

Velocity of the projectile, v = 392 m/s

Angle at which it is fired, $\theta=60^{\circ}$

1. The total horizontal path covered by the projectile is called its range. Its formula is given by :

$R=\dfrac{v^2\ sin2\theta}{g}$

$R=\dfrac{(392)^2\ sin2(60)}{9.8}$

R = 13579.27 meters

2. If H is the maximum height reached by the projectile. Its formula is given by :

$H=\dfrac{v^2\ sin^2\theta}{2g}$

$H=\dfrac{(392)^2\ sin^2(60)}{2\times 9.8}$

H = 5880 meters

So, the range and height of the projectile are 13579.27 m and 5880 m. Hence, this is the required solution.

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Projectile motion

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