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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?

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?

Grade:12

1 Answers

Prajal Gupta
11 Points
2 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)

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