Sir question no 5 mean two Force act kty hain air resistance bhi ha sir aur yahan par mention nh kiya hua ha air resistance ko negligneble to nh kiya hua?
To solve this problem, we need to analyze the vertical motion of the projectile. Given: - Initial speed \( v = 100 \) m/s - Angle with the y-axis \( \theta = 30^\circ \) - Time \( t = 10 \) seconds First, let's find the initial vertical component of the velocity. Since the angle is with the y-axis, the vertical component is given by: \[ v_y = v \cos(\theta) \] Substituting the values: \[ v_y = 100 \cos(30^\circ) \] \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] \[ v_y = 100 \times \frac{\sqrt{3}}{2} \] \[ v_y = 50 \sqrt{3} \text{ m/s} \] Now, we use the kinematic equation for vertical displacement under gravity: \[ y = v_y t - \frac{1}{2} g t^2 \] Where: - \( y \) is the vertical displacement - \( g = 9.8 \) m/s² (acceleration due to gravity) - \( t = 10 \) seconds Substituting the values: \[ y = (50 \sqrt{3}) \times 10 - \frac{1}{2} \times 9.8 \times (10)^2 \] \[ y = 500 \sqrt{3} - 0.5 \times 9.8 \times 100 \] \[ y = 500 \sqrt{3} - 490 \] Calculate \( 500 \sqrt{3} \): \[ 500 \sqrt{3} \approx 500 \times 1.732 = 866 \] So, \[ y = 866 - 490 \] \[ y = 376 \text{ m} \] Therefore, the vertical distance traveled by the projectile in 10 seconds is approximately 376 meters. However, this does not match any of the given options. It seems there might be a mistake in the provided options or the problem statement. Given the closest possible answer based on the options provided, none of them accurately represent the calculated value.
Sir pdf form m y book Hy??
Sir aage bhi mcqs lecture send karn
where is the second part sir
Sir question no 5 mean two Force act kty hain air resistance bhi ha sir aur yahan par mention nh kiya hua ha air resistance ko negligneble to nh kiya hua?
Yes
Sir entry test physics k hawLe se guide krein
q16 ma 1/2 hoga kiyon ki cos inverse ki kabe be 2 nhi hoo sk tha bhale ap calculator kr ka dek hain
Sir plss tell.. Maximum distance kaise find kre ge? Agr sirf initial velocity de ho?
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range max = vi²/g
Range= u² cos theta upon g
R= U²Cos theta/gravity
R=V•²sing2@/g
Sir mere le e book oder kegy please
To solve this problem, we need to analyze the vertical motion of the projectile. Given:
- Initial speed \( v = 100 \) m/s
- Angle with the y-axis \( \theta = 30^\circ \)
- Time \( t = 10 \) seconds
First, let's find the initial vertical component of the velocity. Since the angle is with the y-axis, the vertical component is given by:
\[ v_y = v \cos(\theta) \]
Substituting the values:
\[ v_y = 100 \cos(30^\circ) \]
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
\[ v_y = 100 \times \frac{\sqrt{3}}{2} \]
\[ v_y = 50 \sqrt{3} \text{ m/s} \]
Now, we use the kinematic equation for vertical displacement under gravity:
\[ y = v_y t - \frac{1}{2} g t^2 \]
Where:
- \( y \) is the vertical displacement
- \( g = 9.8 \) m/s² (acceleration due to gravity)
- \( t = 10 \) seconds
Substituting the values:
\[ y = (50 \sqrt{3}) \times 10 - \frac{1}{2} \times 9.8 \times (10)^2 \]
\[ y = 500 \sqrt{3} - 0.5 \times 9.8 \times 100 \]
\[ y = 500 \sqrt{3} - 490 \]
Calculate \( 500 \sqrt{3} \):
\[ 500 \sqrt{3} \approx 500 \times 1.732 = 866 \]
So,
\[ y = 866 - 490 \]
\[ y = 376 \text{ m} \]
Therefore, the vertical distance traveled by the projectile in 10 seconds is approximately 376 meters. However, this does not match any of the given options. It seems there might be a mistake in the provided options or the problem statement. Given the closest possible answer based on the options provided, none of them accurately represent the calculated value.
same point of view
MshaAllah sir g
Option 6 wrong h sir 275 hoga
❤❤❤
Sir ye wali book kaha se milegi? Or kitne ki
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Question 6:
Cos30 1/2 kahan hota hai???
cos 60 ka bataya hy
1/2 tu nae hota
Question ko mention kro
thank you sir
Thnks
Assalam o Alaikum
Walikumsalam
Sir es ka second part nahi mil raha
@@alinamalakMalik upload hai
Sir kindly link ya lecture name bata Dy bohot zarori hai ye lecture
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@@alinamalakMalik projectile motion part 2
Question 6 wrong
What is the name of this book,sir
Physics 5000+ practice mcqs Book by Sayed Umer Farooq.
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@@sayed_umerfarooq tell name of book plz sir
@@cooking_with-MANAQSHA physics 5000+ mcqs ye old addition hai aur new addition 7 days mai ajae ga. Us me 9000+ mcqs hai.
Ali series physics mcqs
@@sayed_umerfarooq 7 days my🥺 hmny tayari krny hyyy test bht qareeb hay
@@cooking_with-MANAQSHA phr ap old book order kre JB new ajae wo b bejo denge
Sir plzzz jwab dydyn
MCQ number 7 is explained wrong
yes answer zero nhi ara
8 mcqs ma 0 teta answer hai