20 Replies
@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.bhai mujhse toh plot nahi hora but kisi graphing calculator par ye plot kar ke dekhna c= (Vx)^2 which is constant with x and C2= net mechanical energy(which is constant too) y= mgh (potential energy as mg is constant and y=h)
and dy/dx= change in vertical velocity ofc 3 ya 4 mein se hi ek hoga
and dy/dx= change in vertical velocity ofc 3 ya 4 mein se hi ek hoga
A is the answer?
A toh ho hi nahi sakta, KE kabhi zero nahi hogi
Potential energy kaha se aagaya bhai?
sorry i meant KE*, uski horizontal velocity is never zero
achaa horizontal displacement hai, right
c?
$$v = ucos\theta \hat{x} + (usin\theta - gt)\hat{y}$$
$$x = ucos\theta t$$
$$t = \frac{x}{ucos\theta}$$
Keshav
Keshav
Keshav
D<0 so, the graph will be (C)
try to find a shorter way to solve this problem
A and B is wrong for sure, cuz KE 0 nhi hogi thorughout the motion
Now C and D mei se dekha jaaye toh its very rare in classical mechanics ki D jaisa non - differentiable (at a point) curve bane
+ projectile motion mei slope mei itna sharp turn nhi milega
achanak se slope nhi badal jaati uski
Therefore, D is more probable to be wrong
which leaves us with C
dy/dt* sorry;(
mene bhi exact same method lagaya hai.... v^2 will be y aur x^2 ka coefficient positive hai toh upward parabola hoga aur ke zero nahi ho sakti so option b cancels toh option c
Yep c is the correct one
Thanx everyone
Got it
I didn't know how to archive then , my bad
+solved @vj_25 @Keshav
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