From top of a tower 160 m high a body is projected upwards with a velocity of 40 m/s making an angle of 30° with the horizontal. Find the time it will take to reach the ground
To find the time it will take for the body to reach the ground, we can use the equations of motion.
Step 1: Resolve the initial velocity into horizontal and vertical components.
The initial velocity can be resolved into its horizontal and vertical components using trigonometry.
Horizontal component: Vx = V * cos(theta)
Vx = 40 m/s * cos(30°)
Vx = 40 m/s * (√3/2)
Vx = 20√3 m/s
Vertical component: Vy = V * sin(theta)
Vy = 40 m/s * sin(30°)
Vy = 40 m/s * (1/2)
Vy = 20 m/s
Step 2: Calculate the time it takes to reach the highest point.
To calculate the time it takes to reach the highest point, we can use the equation:
Vy = Vo + at
We know that at the highest point, Vy = 0 m/s (since the object momentarily stops moving vertically).
0 m/s = 20 m/s - 9.8 m/s^2 * t1
Solving for t1:
t1 = 20 m/s / 9.8 m/s^2
t1 ≈ 2.04 seconds
Step 3: Calculate the time it takes to reach the ground.
To calculate the time it takes to reach the ground, we can use the equation:
h = Vo * t - 0.5 * g * t^2
Since the initial vertical velocity is 20 m/s (Vy), and we want to find the time it takes to reach the ground, we set h = -160 m (negative indicates motion in the downward direction):
-160 m = 20 m/s * t - 0.5 * 9.8 m/s^2 * t^2
This is now a quadratic equation in the form: at^2 + bt + c = 0, where a = -0.5 * 9.8 m/s^2, b = 20 m/s, and c = -160 m.
Solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
t = (-20 ± √(20^2 - 4 * (-0.5 * 9.8) * (-160))) / (2 * (-0.5 * 9.8))
t = (-20 ± √(400 + 3136)) / (-9.8)
t = (-20 ± √3536) / (-9.8)
After solving this equation, we get two solutions: -1.71 seconds and 9.29 seconds. Since time cannot be negative, the object takes approximately 9.29 seconds to reach the ground.
Therefore, the time it will take for the body to reach the ground is approximately 9.29 seconds.
To find the time it will take for the body to reach the ground, we can break down the initial velocity into its vertical and horizontal components.
Given:
Height of the tower (h) = 160 m
Initial velocity (v) = 40 m/s
Angle with the horizontal (θ) = 30°
First, we find the vertical component of the initial velocity:
Vertical component (v_y) = v * sin(θ)
v_y = 40 * sin(30°)
v_y ≈ 20 m/s
Next, we use the equation of motion to calculate the time it will take for the body to reach the ground:
h = v_y * t - (1/2) * g * t^2
where g is the acceleration due to gravity, approximately equal to 9.8 m/s^2.
Rearranging the equation and substituting the values, we get:
0 = -4.9 * t^2 + 20 * t - 160
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
Here, a = -4.9, b = 20, and c = -160.
Let's calculate:
t = (-20 ± sqrt((20)^2 - 4 * (-4.9) * (-160))) / (2 * (-4.9))
t = (-20 ± sqrt(400 + 3136))/ (-9.8)
t = (-20 ± sqrt(3536))/ (-9.8)
Taking the positive root since time cannot be negative:
t = (-20 + sqrt(3536))/ (-9.8)
Calculating the expression inside the square root:
sqrt(3536) ≈ 59.45
Now, let's find the time:
t ≈ (-20 + 59.45)/ (-9.8)
t ≈ 3.08 seconds
Therefore, it will take approximately 3.08 seconds for the body to reach the ground.
initial upward speed is 20 m/s
h(t) = 160 + 20t - 4.9t^2
when is h=0?