Question

A projectile is fired with the initial velocity of 90m/s to hit a ground level target. Its maximum horizontal range will be? ​

Answer 1

Explanation:

Calculating the time taken by the projectile to reach the ground:

s=ut+21gt2

500m=(100m/s)t+21(9.8m/s2)t2

500=100t+4.9t2

t=4.15s

Calculating the distance of the target from the hil:

Range = ut=(100m/s)(4.15s)=415m

Calculating the velocity with which the projectile hits the ground:

v=100m/s+(9.8m/s2)(4.15s)2=268.78m/s

Answer 2

Its maximum horizontal range will be 413.37 m.

Given :

Initial velocity = 90m/s

To find :

Its maximum horizontal range

Solution :

We know that,

For the maximum range,

the angle it makes with horizontal (A) = 45°

According to the formula,

Range = u^2sin2A /g

g is the acceleration due to gravity

g = 9.8m/s^2

Hence, u^2sin2A /g

= 4051.067 × 1 / 9.8

= 413.37 m

Hence, Its maximum horizontal range will be 413.37 m.

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