Question

when a body is projected at an angle to the horizontal,it crosses a point p after 4seconds and then after 5 more seconds it reaches the ground.then height of p above the ground is​

Answer 1

Answer:

R = 40 m   ; it was projected horizontally

So that horizontal velocity =Rt = \frac{R}{t} =tR​

t = time ti free fall a height 20 m

                ↓\downarrow ↓

as initial vertical velocity = 0

Given by h=129t2 h= \frac{1}{2} 9t^{2}h=21​9t2       we have t =2h9 = \sqrt{\frac{2h}{9}}=92h​

​

t=2×2010 t = \sqrt{\frac{2\times 20}{10}} t=102×20​

​ = 2sec

& vertical velocity gained in these 2 sec

vy=u+9t=0+10×2=20 v_{y} = u + 9t = 0 + 10\times 2 = 20 vy​=u+9t=0+10×2=20 m/s  

Horizontal velocity vx=Rt=402=20m/sv_{x}=\frac{R}{t}=\frac{40}{2}= 20 m/svx​=tR​=240​=20m/s (constant)

⇒\Rightarrow ⇒ speed =vy2+vx2=202 = \sqrt{v^{2}_{y}+v^{2}_{x}} = 20\sqrt{2} =vy2​+vx2​

​=202

​ m/s

solution

Explanation:

Answer 2

Answer:

Given:

A Projectile crosses point p after 4 seconds. After 5 more seconds, it reaches the ground.

To find:

Height of point p above ground.

Calculation:

So total time for Projectile to reach ground = 4 + 5 = 9 secs.

Let angle of Projection be θ

$\therefore \: \dfrac{2u \sin( \theta) }{g} = 9$

$= > u \: \sin( \theta) = \dfrac{9g}{2} = 45 \: m {s}^{ - 1}$

Now height of p above ground:

$h = u \sin( \theta)\times t \: - \frac{1}{2} g {t}^{2}$

$= > h = (45 \times 4) - ( \frac{1}{2} \times 10 \times {4}^{2} )$

$= > h = 180 - 80$

$= > h = 100 \: metres$

So final answer is :

$\boxed{ \red{height = 100 \: metres}}$