Question 26: Motion in 1D

Question

A ball is thrown vertically upward with an initial velocity of $$150 \mathrm{~m} / \mathrm{s}$$. The ratio of velocity after $$3 \mathrm{~s}$$ and $$5 \mathrm{~s}$$ is $$\frac{x+1}{x}$$. The value of $$x$$ is ___________.
$$\left\{\right.$$ take, $$\left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right\}$$

Accepted Answer

To solve this problem, we can use the following equation of motion for the vertical velocity at any given time $$t$$:
$$v = u - gt$$
Where:
- $$v$$ is the final velocity at time $$t$$
- $$u$$ is the initial velocity (150 m/s)
- $$g$$ is the acceleration due to gravity (10 m/sÃÂ²)
- $$t$$ is the time in seconds
First, we need to find the velocities at $$t = 3 \mathrm{~s}$$ and $$t = 5 \mathrm{~s}$$.
For $$t = 3 \mathrm{~s}$$:
$$v_3 = 150 - (10)(3) = 150 - 30 = 120 \mathrm{~m} / \mathrm{s}$$
For $$t = 5 \mathrm{~s}$$:
$$v_5 = 150 - (10)(5) = 150 - 50 = 100 \mathrm{~m} / \mathrm{s}$$
Now we need to find the ratio of these velocities:
$$\frac{v_3}{v_5} = \frac{120}{100} = \frac{6}{5} = \frac{x + 1}{x}$$
Next, we can set up an equation to find the value of $$x$$:
$$\frac{6}{5} = \frac{x + 1}{x}$$
Now, we can solve for $$x$$:
$$6x = 5(x + 1)$$
$$6x = 5x + 5$$
$$x = 5$$
The value of $$x$$ is 5.

Answer

$$-5$$

Answer

10

Answer

5

Answer

6

Motion in 1D

A ball is thrown vertically upward with an initial velocity of $150 \mathrm{~m} / \mathrm{s}. The ratio of velocity after 3 \mathrm{~s} and 5 \mathrm{~s} is \frac{x+1}{x}. The value of x is ___________. \left\{\right. take, \left.g=10 \mathrm{~m} / \mathrm{s}^{2}\right\}

Choose the correct answer: