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Grade 12Electric Current

A missile was fired with a velocity of 30m/s making an angle 60' with the horizontal. It strikes a hill which is at a horizontal distance of 45m from the point o projection. What is the height above the ground level; the point where the missile strike the hill?

Aravind

7 Years agoGrade 12

1 Answer

Prajal Gupta

7 Years ago

At any horizontal distance x, the corresponding height of the body in a projectile motion is given by

y = xtanA - (x/ucosA) ² g/2

Where A is angle of projection with the horizontal.

Here missile is projected at an angle of 60° with the horizontal. And it travels a horizontal distance of 45 m. We have to calculate the height above the ground corresponding to the point on ground ( foot of hill).

So, here x = 45 m, A = 60°, u = 30 m/s and g= 10 m/s²

Putting these values in the above formula - we get

y = 45 × tan60° - 5(45.2/30)²

= 45(√3 - 1)

The fig. Shows a network of resistance in which the point A is earthed

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