Question

A particle is thrown with a speed of 12 m/s at an angle 60° with the horizontal. The time
interval between the moments when its speed is 10 m/s is (g = 10 m/s2)​

Answer 1

Explanation:

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Answer 2

Answer:

Horizontal velocity $v_{h}$ =u cos∅

Vertical velocity $v_{v}$ =$\sqrt{v^{2} +v_{h} ^{2} }$

Explanation:

Given , a partical is throw with a speed of 12m/s at an angle 60° with

horizontal .

we know that , Horizontal velocity $v_{h}$ =u cos∅ and Vertical velocity $v_{v}$ =$\sqrt{v^{2} +v_{h} ^{2} }$

u = 12m/s   (initial velocity)

v=10 m/s     (final velocity)

∅ = 60°     (angle)

t is the time .

Step 1:

$v_{h}$ =u cos∅ ................(i)

put the value of  u = 12m/sand ∅  = 60°

we get ,$v_{h}$ =12 ×cos60°

⇒$v_{h}$ =12 × $\frac{1}{2}$ = 6

∴$v_{h}$ =6m/s

Step 2:

we already mention that ,

vertical velocity  $v_{v}$ =$\sqrt{v^{2} +v_{h} ^{2} }$ ...................(ii)

From (step 1 )we get  $v_{h}$= 6m/s  in the question

and  v = 10 m/s given in the question

Now,  put the value of  v and $v_{h}$ in equation (ii)

so ,$v_{v}$ = $\sqrt{10^{2} -6^{2} }$

⇒$v_{v}$ =$\sqrt{100-36} = \sqrt{64}$=8

$v_{v}$ = 8 m/s

Step 3:

$t_{1}$ = (usin∅ - $v_{v}$)/v and $t_{2}$ = (usin∅ + $v_{v}$)/v

[where $t_{1}$  is the time taken at speed 12m/s and $t_{2}$   is the time taken   when speed is 10m/s]

∴ $t_{1}$ = (usin∅ - $v_{v}$)/v  = (usin∅ - 8)/10

and $t_{2}$ = (usin∅ + $v_{v}$)/v = (usin∅ + 8)/10

Now , time interval is

$t_{2}$ -$t_{1}$ =  (usin∅ + $v_{v}$)/v  - (usin∅ - $v_{v}$)/v  ..............(iii)

put all the values in the equation (iii)

$t_{2}$ -$t_{1}$ = $\frac{8+8}{10}$

⇒  $t_{2}$ -$t_{1}$ =1.6s

Final answer :

Hence , the time interval between the moments when its speed 10m/s is

1.6sec