Question

particle is projected horizontally from the top of a building of height 20m with a speed of 30 m s . distance of point where particle lands on ground from its point of projection is

a. 600 m
b. 5√10 m
c. 20√10 m
d. 30√10 m ​

Answer 1

Explanation:

R = 40 m   ; it was projected horizontally

So that horizontal velocity =tR

t = time ti free fall a height 20 m

                 ↓

as initial vertical velocity = 0

Given by h=219t2       we have t =92h

t=102×20 = 2sec

& vertical velocity gained in these 2 sec

vy=u+9t=0+10×2=20 m/s  

Horizontal velocity vx=tR=240=20m/s (constant)

⇒ speed =vy2+vx2=202 m/s

Hope this answer is very helpful

Answer 2

Given:

Particle is projected horizontally from the top of a building of height 20m with a speed of 30 m/s.

To find:

Distance between landing point and point of Projection?

Calculation:

Let the time taken to reach the ground be "t":

$\therefore \: H = u_{y}t + \dfrac{1}{2} g {t}^{2}$

$\implies\: H = (0)t + \dfrac{1}{2} g {t}^{2}$

$\implies\: H = \dfrac{1}{2} g {t}^{2}$

$\implies\:20 = \dfrac{1}{2} (10){t}^{2}$

$\implies\:{t}^{2} = 4$

$\implies\:t = 2 \: sec$

Now , let horizantal distance travelled be R :

$\therefore \: R = u_{x} \times t$

$\implies \: R = 30 \times 2$

$\implies \: R = 60 \: metres$

So, the required distance be D :

$\therefore D = \sqrt{R^{2}+H^{2}}$

$\implies D = \sqrt{{60}^{2}+{20}^{2}}$

$\implies D = \sqrt{3600+400}$

$\implies D = \sqrt{4000}$

$\implies D = \sqrt{400\times 10}$

$\implies D = 20\sqrt{10} \:metres$

So, required distance is 20√10 m.