Projectile Motion

Projectile motion refers to the motion of an object that is launched into the air and continues to move under the force of gravity. It follows a curved path known as a parabola. In this case, we are given that the maximum height reached by the projectile is 4m and the horizontal range is 12m. We can use these values to determine the velocity of projection.

Maximum Height

The maximum height reached by a projectile occurs when its vertical velocity becomes zero. At this point, the object starts to fall back down. Let's assume the initial velocity of projection is v.

Using the equations of motion, we can determine the time taken for the projectile to reach its maximum height:

v = u + at

Since the vertical velocity becomes zero at maximum height, the final velocity (v) is zero. The initial velocity (u) is unknown. The acceleration (a) is -9.8 m/s^2 (acceleration due to gravity).

0 = u - 9.8t

Solving for t, we get:

t = u/9.8

The time taken to reach maximum height is the same as the time taken to fall back down to the ground, so the total time of flight is 2t.

The maximum height (H) is given by:

H = ut + (1/2)at^2

Substituting the value of t, we get:

H = (u^2)/(9.8) - (u^2)/(2*9.8)

Simplifying further, we find:

H = (u^2)/(2*9.8) [Equation 1]

Given that H = 4m, we can solve Equation 1 for u:

4 = (u^2)/(2*9.8)

u^2 = 4 * 2 * 9.8

u^2 = 78.4

u = √78.4

Velocity of Projection

The velocity of projection (v) is given by:

v = u + at

Substituting the known values, we have:

v = √78.4 - 9.8t

Since the time taken to reach maximum height is t = u/9.8, we can substitute this value:

v = √78.4 - 9.8(u/9.8)

v = √78.4 - u

v = √78.4 - √78.4

v = 0

Therefore, the velocity of projection is zero. This means that the projectile was launched vertically upwards and then fell back down under the force of gravity.

Horizontal Range

The horizontal range (R) is the distance covered by the projectile horizontally. It is given by:

R = ut

Substituting the known values, we have:

12 = u(t)

12 = √78.4(t)

t = 12/√78.4

t ≈ 1.36 seconds

Therefore, the projectile takes approximately 1.36 seconds to reach the ground.

Conclusion

In summary, the given information of the maximum height reached by the projectile being 4m and the horizontal range being 12m allows us to determine the velocity of projection. By solving the equations of motion and using the known