To determine the velocity of projection, we can use the fact that the time of flight of a projectile is twice the time it takes for the projectile to reach its maximum height. We'll also assume that there is no air resistance.
Let's denote the velocity of projection as 'v' and the time it takes for the projectile to reach its maximum height as 't'. The time of flight is then 2t.
Since the projectile reaches a maximum height and returns to the same horizontal plane, the vertical component of velocity at the highest point is zero.
Using the equations of motion for the vertical direction:
At the highest point: v_y = 0 v_y = u_y + a_y * t 0 = v * sinθ - g * t t = v * sinθ / g
Where: v_y is the vertical component of velocity u_y is the initial vertical component of velocity (which is v * sinθ) a_y is the acceleration due to gravity (-g) t is the time it takes to reach maximum height
Now, considering the horizontal motion, we know that the horizontal displacement is 60 meters and the time of flight is 4 seconds:
x = u_x * t 60 = v * cosθ * (2t)
Since t = v * sinθ / g, we can substitute this expression into the equation:
60 = v * cosθ * (2 * (v * sinθ / g))
Simplifying the equation:
60 = (2v² * sinθ * cosθ) / g
Using the trigonometric identity sin(2θ) = 2sinθ * cosθ, we can rewrite the equation as:
60 = (v² * sin(2θ)) / g
Now we can solve for v:
v² = (60 * g) / sin(2θ) v = √((60 * g) / sin(2θ))
Finally, we need to find the angle θ. Since the vertical component of velocity at the highest point is zero, we know that the time taken to reach the highest point is half of the total time of flight. Thus:
t = 4 seconds / 2 = 2 seconds
Now we can substitute this time into the equation for t above:
2 = v * sinθ / g
Simplifying:
sinθ = (2 * g) / v θ = arcsin((2 * g) / v)
Substituting the expression for v from the previous equation, we get:
θ = arcsin((2 * g) / √((60 * g) / sin(2θ)))
Solving for θ numerically using an equation solver would yield the angle of projection.