Question

The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be

Answer 1

Explanation:

Let initial velocity = u m/sec

Range = 400 m

Maximum Height =?

As we know that

For maximum Range, Angle of projection (A) = 45°

Range = u^2 sin2A /g

400 = u^2 ×sin2×45°/g

400 = u^2 sin90°/g

400 = u^2 /g - - - - - - (1)

Maximum Height = u^2 sin^2 A/2g

= u^2 /g × sin^245°/2

= 400 × 1/2/2

= 400 × 1/4

= 100 m

-—————-—-———-—------------¦

Maximum Height = 100 m¦

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Answer 2

Answer: H=100m

Explanation:

For the maximum range, we throw the object at 45° angle.

From formula of,

R=u^2 sin(2y)/g

H=u^2 sin^2(y)/2g

Where u is initial velocity, y is angle the ball was thrown at, g is gravitational acceleration.

Thus by solving the above equations, we get

=> R/H=4/tany

=> R=4*H *coty

=> 400/(4*1)=H

=> H=100m

Thus the maximum height of the projectile is 100m.