A projectile is fired with an initial speed of 32.5 m/s at an angle of 61.4° above the horizontal. The object hits the ground 7.55 s later. What is the absolute value of the difference in height between the launch point and the point where the projectile hits the ground?
the vertical point of the projectile at any time t is
h(t)=32.5Sin61.4 * t - 4.9 t^2
To find the absolute difference in height between the launch point and the point where the projectile hits the ground, we can use the equations of motion for projectile motion.
First, we need to determine the initial vertical velocity (Vy) and the time of flight (t).
Given that the initial speed (V₀) is 32.5 m/s and the launch angle (θ) is 61.4° above the horizontal, we can find Vy using trigonometry:
Vy = V₀ * sin(θ)
Vy = 32.5 * sin(61.4°)
Vy ≈ 28.13 m/s
Next, we can find the time of flight (t) using the vertical motion equation:
Δy = Vy * t + (1/2) * g * t²
Since the projectile returns to the same height at the launch point, Δy = 0. Therefore, we can solve for t:
0 = Vy * t + (1/2) * g * t²
Using the value of gravitational acceleration (g ≈ 9.8 m/s²), we can solve this quadratic equation to find t, which will be the time of flight:
0 = (1/2) * 9.8 * t² + 28.13 * t
Solving this equation gives two solutions: t = 0s and t ≈ 5.12s. Since the time cannot be zero, we take t ≈ 5.12s as the time of flight.
Now, to find the difference in height, we can use the formula:
Δy = Vy * t + (1/2) * g * t²
Substituting the values we have:
Δy = 28.13 * 5.12 + (1/2) * 9.8 * (5.12)²
Δy ≈ 143.66 m
Therefore, the absolute value of the difference in height between the launch point and the point where the projectile hits the ground is approximately 143.66 meters.