A person standing on a tower of height 60 m throws an object upwards with velocity of 40 m/s at an angle 30° to horizontal. Find the total time taken by the object to gain maximum height and fall on the ground (take g = 10 m/s2)
6 s
Object Thrown Upwards: Total Time Calculation
This problem involves projectile motion from a certain height, where we need to find the total time the object remains in the air until it hits the ground. We will analyze the vertical motion of the object using kinematic equations.
Time Calculation: Understanding Projectile Motion
Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. To find the total time the object takes to reach the ground, we focus on its vertical displacement, initial vertical velocity, and the acceleration due to gravity.
Given Parameters: Tower and Object Details
Let's list the information provided in the question:
- Height of the tower (\(h_0\)) = 60 m
- Initial velocity of the object (\(u\)) = 40 m/s
- Angle of projection (\(\theta\)) = 30° to the horizontal
- Acceleration due to gravity (\(g\)) = 10 m/s2
We need to find the total time (\(t\)) taken by the object to reach its maximum height and then fall to the ground.
Vertical Motion: Analyzing Object's Trajectory
First, we need to determine the initial vertical component of the object's velocity, as this is crucial for analyzing its vertical motion.
- Initial Vertical Velocity (\(u_y\)):
The vertical component of the initial velocity is given by \(u_y = u \sin \theta\). \[u_y = 40 \sin 30^\circ\] Since \(\sin 30^\circ = \frac{1}{2}\): \[u_y = 40 \times \frac{1}{2} = 20 \text{ m/s}\] - Vertical Displacement (\(s_y\)):
The object starts from a height of 60 m and eventually falls to the ground (0 m height). Therefore, the total vertical displacement from its starting point (top of the tower) to the ground is -60 m (negative because the final position is below the initial position). \[s_y = -60 \text{ m}\] - Acceleration due to gravity (\(a_y\)):
Gravity always acts downwards. If we consider the upward direction as positive, then the acceleration due to gravity is negative. \[a_y = -g = -10 \text{ m/s}^2\]
Applying Kinematic Equation for Total Time
We will use the second kinematic equation, which relates displacement, initial velocity, acceleration, and time:
\[s_y = u_y t + \frac{1}{2} a_y t^2\]Now, substitute the values we have into the equation:
\[-60 = (20)t + \frac{1}{2} (-10) t^2\] \[-60 = 20t - 5t^2\]Rearrange this equation into a standard quadratic form \(at^2 + bt + c = 0\):
\[5t^2 - 20t - 60 = 0\]To simplify, divide the entire equation by 5:
\[t^2 - 4t - 12 = 0\]Solving the Quadratic Equation for Total Time
We can solve this quadratic equation by factorization. We need two numbers that multiply to -12 and add up to -4. These numbers are -6 and +2.
\[(t - 6)(t + 2) = 0\]This gives two possible solutions for \(t\):
- \(t - 6 = 0 \implies t = 6 \text{ s}\)
- \(t + 2 = 0 \implies t = -2 \text{ s}\)
Since time cannot be a negative value, we discard \(t = -2 \text{ s}\).
Therefore, the total time taken by the object to reach its maximum height and then fall on the ground is 6 seconds.
Important Questions from Projectiles
The range of a projectile is maximum, when the angle of projection is -
'If 'θ' is angle of projection and 'u' is velocity of projection for a projectile, then its horizontal range is given by-
A particle is projected with velocity u at an inclination θ with the horizontal. Then Maximum height (H) attained is
The angle of projection of a projectile is _________ when the range will be maximum for a given velocity.
Which of the following is NOT a projectile motion?
