Vertical Motion Displacement Calculation
To find the displacement of the ball after 4 seconds, we use the kinematic equation for displacement under constant acceleration:
\(s = ut + \frac{1}{2}at^2\)
Where:
- \(s\) is the displacement
- \(u\) is the initial velocity
- \(t\) is the time
- \(a\) is the acceleration
Applying Values and Calculation
Given:
- Initial velocity, \(u = 30\ m/s\) (positive as it's upward)
- Time, \(t = 4\ s\)
- Acceleration due to gravity, \(g = 10\ m/s^2\). Since gravity acts downwards, opposing the initial upward motion, the acceleration \(a = -g = -10\ m/s^2\).
Substitute these values into the equation:
\(s = (30\ m/s)(4\ s) + \frac{1}{2}(-10\ m/s^2)(4\ s)^2\)
First, calculate the terms:
- \(ut = (30)(4) = 120\ m\)
- \(\frac{1}{2}at^2 = \frac{1}{2}(-10)(16) = -5 \times 16 = -80\ m\)
Now, find the displacement:
\(s = 120\ m - 80\ m\)
\(s = 40\ m\)
The magnitude of the displacement is 40 m.
Similar Questions
- A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
- A ball is thrown vertically upward with a speed of 30 m/s. The magnitude of its displacement after 4 s will be ______ (Take \(g = 10 \text{ m/s}^2\).)
- A body falling from rest has a velocity '\(v\)' after it falls through a distance '\(h\)'. The distance it has to fall down further, for its velocity to become double, is _____ times '\(h\)' .
- By applying the brakes without causing a skid, the driver of a car is able to stop his car within a distance of 5 m, if it is going at 36 kmph. If the car were going at 72 kmph, using the same brakes, he can stop the car over a distance of:
- A ball, initially at rest, falls freely from the top of a building and reaches a maximum velocity of \(40\text{ m/s}\). Find the height of the building. (Use acceleration due to gravity, \(g = 10\text{ m/s}^2\))
- A car, starting from rest, is moving with a constant acceleration \(3\text{ m/s}^2\). Find the distance travelled by this car in \(10\text{ s}\).
- A ball is thrown vertically upwards with initial velocity \(V_0\) and returns to its starting point in 6 seconds then the initial velocity with which the ball was thrown will be
- A ball, having speed \(V_0\) moves in a straight line under the influence of a constant acceleration a. Then its final speed after travelling a distance x for time t will be
A particle experiences constant acceleration for 20 s after starting from rest. If it travels a distance X 1, in the first 10 s and distance X 2in the remaining 10 s, then which of the following is true?
Important Questions from Kinematic equations for uniformly accelerated motion
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
- A ball is thrown vertically upward with a speed of 40 m/s. The time taken by the ball to reach the maximum height would be approximately
- A ball thrown up vertically returns to the ground after 10 second. Find the velocity with which it was thrown up? (if g = 10 m/s2).
If the distance travelled by a body in the $n^{th}$ second is given by $(7 + 5n)$ m, then find the initial velocity and acceleration of the body respectively.
Which of the following is an equation of motion?
I. u = v + at
II. 2as = v 2– u 2
