Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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?

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)