what maximum height will a 311.2 m drive reach if it is launched at an angle of 17.0 degrees to the ground
To find the maximum height reached by a projectile launched at an angle, we can use the equations of projectile motion. The maximum height is the vertical displacement when the vertical velocity component becomes zero.
To start, we need to find the initial vertical velocity (Vy) and the initial horizontal velocity (Vx). We can use the given angle and the total initial velocity (V) to calculate these components.
Vy = V * sin(angle)
Vx = V * cos(angle)
Next, we need to find the time taken for the projectile to reach its maximum height. We can use the equation of motion:
Vy = V0y + (-g) * t
where Vy is the final vertical velocity, V0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s²), and t is the time taken.
Since the final vertical velocity at the maximum height is zero, the equation becomes:
0 = Vy + (-g) * t
We can now solve this equation for t:
t = Vy / g
With the time taken (t), we can find the maximum height (H) using the equation of motion:
H = V0y * t + (1/2) * (-g) * t²
Substituting the value of t, we get:
H = Vy² / (2 * g)
Now, let's plug in the values given in your question:
V = 311.2 m/s
angle = 17.0 degrees
g = 9.8 m/s²
Calculating Vx and Vy:
Vx = V * cos(angle) = 311.2 * cos(17.0) = 298.07 m/s
Vy = V * sin(angle) = 311.2 * sin(17.0) = 86.05 m/s
Calculating the time taken:
t = Vy / g = 86.05 / 9.8 = 8.79 s
Calculating the maximum height:
H = Vy² / (2 * g) = (86.05)² / (2 * 9.8) = 390.61 m
Therefore, the maximum height reached by the projectile will be approximately 390.61 meters.