Question

173. If time of flight of a projectile is 10 seconds. Range
is 500 meters. The maximum height attained by it
will be :-
(1) 125 m
(2) 50 m
(3) 100 m
(4) 150 m​

Answer 1

Given:

• Time of flight = 10s
• Range = 500 m

To find:

• Maximum height

Solution:

As we know that

T = 2usinθ/g

Substituting g = 10 m/s² beacause acceleration due to gravity is 9.8 m/s² which is approximately ≈ 10 m/s²

→ 10 = 2usinθ/10

→ 10 × 10 = 2usinθ

→ 100/2 = usinθ

→ usinθ = 50 m/s

Squaring on both sides

→ u²sin²θ = 2500

Assuming as equation 1

Now , using

Maximum height reached = u²sin²θ/2g

→ Max. h reached = 2500/2(10). [°.° Eq. 1 ]

→ Max. height reached = 2500/20

→ Max. height reached = 125 m

Hence , option 1 is your answer