Question

A body of mass 5kg projected at an angle of 60∘ from the ground with an initial velocity of 25m/s, acceleration due to gravity is g=10m/S2, what is the maximum horizontal range covered?

Answer 1

To find the maximum horizontal range covered by a projectile, we can use the following formula:

Range = (initial velocity squared * sin(2*theta)) / gravity

where:

• initial velocity is the magnitude of the initial velocity (25 m/s)
• theta is the angle of projection (60 degrees)
• gravity is the acceleration due to gravity (10 m/s^2)

Let's calculate the maximum horizontal range:

theta = 60 degrees initial velocity = 25 m/s gravity = 10 m/s^2

First, we need to convert the angle from degrees to radians:

theta_rad = theta * (pi/180) theta_rad = 60 * (pi/180) theta_rad = pi/3

Now, we can substitute the values into the formula:

Range = (25^2 * sin(2*(pi/3))) / 10

Calculating sin(2*(pi/3)): sin(2*(pi/3)) = sin(2*(pi/3)) = sin(4*(pi/6)) = sin(4*(pi/3)) = sin(2*pi) = 0

Therefore, the range becomes: Range = (25^2 * 0) / 10 Range = 0

Hence, the maximum horizontal range covered by the projectile is 0 meters.