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
PREPARATORY SCHOOL GRADE 11
To answer the given questions, we will use the equations of projectile motion. Let's break down each question and solve it step by step:
(a) To find the position of the projectile at time t=2s, we need to calculate the horizontal and vertical components separately.
The horizontal component of the projectile's position is given by the equation:
dx = v * cos(θ) * t
where:
dx = horizontal position
v = initial velocity = 30 m/s
θ = launch angle = 60 degrees
t = time = 2s
Plugging in the values:
dx = 30 * cos(60) * 2
dx = 30 * 0.5 * 2
dx = 30
So, the horizontal position of the projectile at t=2s is 30 meters.
The vertical component of the projectile's position is given by the equation:
dy = v * sin(θ) * t + (1/2) * g * t^2
where:
dy = vertical position
v = initial velocity = 30 m/s
θ = launch angle = 60 degrees
t = time = 2s
g = acceleration due to gravity = 9.8 m/s^2
Plugging in the values:
dy = 30 * sin(60) * 2 + (1/2) * 9.8 * 2^2
dy = 30 * (√3/2) * 2 + (1/2) * 9.8 * 4
dy = 30 * √3 + 9.8 * 2
dy ≈ 51.96
So, the vertical position of the projectile at t=2s is approximately 51.96 meters.
The magnitude of the velocity at t=2s can be calculated using the formula:
v = √(v_x^2 + v_y^2)
where:
v = magnitude of velocity
v_x = horizontal component of velocity
v_y = vertical component of velocity
The horizontal component of velocity remains constant throughout the motion:
v_x = v * cos(θ) = 30 * cos(60) = 30 * 0.5 = 15 m/s
The vertical component of velocity changes due to gravity:
v_y = v * sin(θ) - g * t = 30 * sin(60) - 9.8 * 2 ≈ 25.98 m/s
Now we can calculate the magnitude of the velocity:
v = √(15^2 + 25.98^2) ≈ 30.3 m/s
The direction of velocity can be found using trigonometry:
θ_v = arctan(v_y / v_x) = arctan(25.98 / 15) ≈ 59.4 degrees
Therefore, at t=2s, the magnitude of the velocity is approximately 30.3 m/s, and its direction is approximately 59.4 degrees.
(b) To find the time when the projectile reaches its highest point, we can use the fact that the vertical component of velocity becomes 0 at the highest point.
The vertical component of velocity is given by the equation:
v_y = v * sin(θ) - g * t
To find the time at the highest point, we set v_y = 0:
0 = v * sin(θ) - g * t_highest
Rearranging the equation, we get:
t_highest = v * sin(θ) / g
Plugging in the values:
t_highest = 30 * sin(60) / 9.8 ≈ 1.82s
So, the projectile reaches its highest point at approximately t = 1.82 seconds.
(c) To find the horizontal range, we need to calculate the time of flight. The time of flight is the total time taken for the projectile to return to the ground.
The time of flight can be calculated using the formula:
t_flight = 2 * t_highest
Using the value of t_highest from the previous step:
t_flight = 2 * 1.82 ≈ 3.64s
Now, we can find the horizontal range using the formula:
range = v_x * t_flight
Plugging in the values:
range = 15 * 3.64 ≈ 54.6 meters
Therefore, the horizontal range of the projectile is approximately 54.6 meters.