Question

An object is thrown horizontally from a height of 20 m with velocity 10ms-1.Find its velocity after 1s
(g=10ms-2)

Answer 1

An object is thrown horizontally from a height of $20\mathrm{m}$ with velocity 10ms-1. Find its velocity after 1s

$(g = 10\mathrm{ms} - 2)$

Solution:

Resulting velocity after 1s - is the vector sum of velocities along the X-axis and Y-axis:

$$\vec {V} = \vec {V} _ {\mathrm {x}} + \vec {V} _ {\mathrm {y}}$$

Along the X axis there is no acceleration, hence:

$$V _ {x} = V _ {0} = 1 0 \frac {m}{s}$$

Along the Y axis there is acceleration $g = 9.8 \frac{m}{s^2}$, so we can write rate equation for Y axis:

$$V _ {y} = 0 + g t$$ $$V _ {y} = g t = 1 0 \frac {m}{s ^ {2}} \cdot 1 s = 1 0 \frac {m}{s}$$

Now we can find resulting velocity after 1 s (from the right triangle ABC):

$$V = \sqrt {V _ {x} ^ {2} + V _ {y} ^ {2}} = \sqrt {\left(1 0 \frac {m}{s}\right) ^ {2} + \left(1 0 \frac {m}{s}\right) ^ {2}} = 1 0 \sqrt {2} \frac {m}{s} = 1 4 \frac {m}{s}$$

Answer: velocity after 1s will be $14\frac{\mathrm{m}}{\mathrm{s}}$.