An object fired at an angle of 31.9° above the horizontal takes 2.55 s to travel the last 15.5 m of its vertical distance and the last 10.5 m of its horizontal distance. With what speed was the object launched? (Note: This problem does not stipulate that the initial and final elevation of the object has to be the same!)
To find the initial speed at which the object was launched, we can use the equations of motion for projectile motion. Let's break down the problem into horizontal and vertical components.
Horizontal Motion:
We are given that the object travels the last 10.5 m horizontally in 2.55 s. We can use the equation for horizontal distance:
Horizontal distance (d) = initial horizontal velocity (Vx) * time (t)
Vertical Motion:
We are given that the object travels the last 15.5 m vertically in 2.55 s. The equation for vertical distance is more complex since it involves both initial vertical velocity (Vy) and acceleration due to gravity (g). However, since the object is at its maximum height, the vertical velocity at this point is 0 m/s.
Using the equations of motion, we can find the vertical velocity at the maximum height:
Vertical distance (d) = initial vertical velocity (Vy) * time (t) + 0.5 * acceleration due to gravity (g) * time (t)^2
We know the values for the vertical distance (15.5 m), time (2.55 s), and acceleration due to gravity (9.8 m/s^2). We can solve this equation to find the initial vertical velocity (Vy).
Next, we can use trigonometry to determine the vertical and horizontal components of the object's initial velocity. The angle of launch (31.9°) gives us the relationship between the two components:
tan(angle) = vertical component / horizontal component
Finally, we can use the Pythagorean theorem to find the magnitude of the object's initial velocity:
Initial speed = sqrt(vertical component^2 + horizontal component^2)
Using these steps, we can calculate the initial speed at which the object was launched.