A projectile is launched with 30m/s at an angle of 60degree with d horizontal (a) find d position of the projectile,and d magnitude and direction of its velocity at t=2s (b) find d time when d ball reaches the highest point (c) find d horizontal range
To solve this problem, we can break it down into three parts:
(a) Finding the position and velocity at t=2s:
1. Start by decomposing the initial velocity into its horizontal and vertical components. The horizontal component can be found using the formula: v_x = v * cos(theta), where v is the initial velocity and theta is the launch angle.
In this case, v = 30 m/s and theta = 60 degrees. So we have v_x = 30 * cos(60) = 15 m/s.
2. The vertical component can be found using the formula: v_y = v * sin(theta).
In this case, v_y = 30 * sin(60) = 25.98 m/s.
3. To find the position of the projectile at t=2s, we need to find the horizontal distance traveled. We can use the formula: d = v_x * t, where d is the horizontal distance and t is the time.
In this case, d = 15 m/s * 2s = 30 m.
4. To find the magnitude of the velocity at t=2s, we can use the Pythagorean theorem: v = √(v_x^2 + v_y^2).
In this case, v = √(15^2 + 25.98^2) = 30.25 m/s. The direction of the velocity can be found using the inverse tangent function: θ = arctan(v_y / v_x).
In this case, θ = arctan(25.98 / 15) = 59.04 degrees. Thus, the magnitude of the velocity is 30.25 m/s, and the direction is 59.04 degrees (measured from the horizontal axis).
(b) Finding the time when the projectile reaches the highest point:
1. At the highest point, the vertical velocity becomes zero. We can use the equation: v_y = v_0y - g * t, where v_y is the vertical velocity, v_0y is the initial vertical velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
In this case, v_0y = 30 * sin(60) = 25.98 m/s.
2. Set v_y to zero and solve for t: 0 = 25.98 - 9.8 * t.
Rearranging the equation gives t = 25.98 / 9.8 ≈ 2.65 s. So the time when the projectile reaches the highest point is approximately 2.65 seconds.
(c) Finding the horizontal range:
1. The horizontal range is the total horizontal distance traveled by the projectile. To find it, we need to determine the time of flight, which is the total time the projectile is in the air.
The time of flight can be found using the formula: T = 2 * t_max, where T is the time of flight and t_max is the time when the projectile reaches its highest point.
In this case, T = 2 * 2.65 ≈ 5.3 s.
2. The horizontal range can be determined using the equation: R = v_x * T, where R is the horizontal range and v_x is the horizontal component of the initial velocity.
In this case, R = 15 m/s * 5.3 s ≈ 79.5 m. So the horizontal range is approximately 79.5 meters.