A projectile is fired from the surface of the earth with an initial speed of 40 ~m/s . If its speed at the highest point of its trajectory is exactly half of its initial projection speed, what is the maximum height attained by the projectile? (Take g = 10 ~m/s^2 and neglect air resistance)

Step-by-step solution

Let the initial speed be u and the angle of projection with the horizontal be . At the highest point of the trajectory, the vertical component of velocity becomes zero, and the projectile only has a horizontal component of velocity, which remains constant throughout the motion. Thus, the speed at the highest point is v = u . Given that the speed at the highest point is half of the initial speed: u = u 2 = 1 2 = 60^ The maximum height H is given by the formula: H = u^2 ^2 2g Substituting u = 40 ~m/s , = 60^ , and g