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Étiquette : GRADE 11 PHYSICS
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Rocket Vertical Acceleration: Calculating Time to Reach Maximum Point
Question: A rocket is fired vertically up from the ground with a resultant vertical acceleration of 15 ms-2.The fuel is finished in 1 minute and it continues to move up.What is the time taken for the rocket to reach its maximum point? (take g =10 ms-2 ) Solution: Vertical acceleration- calculating time to reach maximum…
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Ratio of Time Taken to Drop Bodies from Different Heights
Question: Two bodies of different masses m1 and m2 are dropped from two different heights a and b.What is the ratio of time taken by the two bodies to drop through these distances. Solution: Ratio of time taken The two objects are striking the ground with velocities v1 and v2 respectively. $${v_1=\sqrt{2gh_1}}$$ $${v_2=\sqrt{2gh_2}}$$ Time taken…
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Projectile Motion: Angle for Equal Height and Range
Question What is the value of angle of projection for which the height of the projectile (H) and the range (R)of the projectile are the same? Solution: Height and range Maximum height of the projectile (H) Height of the projectile (H) $${H=\frac{u^2 \sin^2\theta}{2g}}$$ Range of the projectile $${R=\frac{u^2\sin 2 \theta}{g}}$$ $${H=R}$$ $${\frac{u^2\sin^2\theta}{2g}=\frac{u^2\sin 2\theta}{g}}$$ $${\frac{\sin^2\theta}{2}=2\sin\theta\cos\theta}$$ $${\frac{\sin\theta}{\cos\theta}=4}$$…
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Inclined Plane: Resolving Weight Forces
Question: A mass of 2 Kg lies on an inclined plane as shown below.Resolve its weight along and perpendicular to the plane. (take g=10 ms-2 ) Solution: Inclined plane shopify.pxf.io/1r4onx Resolving the weight into two components The component which is along the plane is $${mg\sin\theta}$$ $${20 \sin\ 30^\circ}$$ $${20 (\frac{1}{2})}$$ $${=10 \ N}$$ The component…
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Anti-Aircraft Shell Vertical Ascent: Calculations and Velocities
Question: An anti-aircraft shell is fired vertically upwards with a muzzle velocity of 294 ms-1.Calculate Solution: (1) Anti-aircraft shell-Maximum height Condition for maximum height(H) is $${\sin^2\theta=1}$$ $${H=\frac{u^2}{2g}=\frac{294^2}{2(9.8)}}$$ $${H=(15)(294)=4410 \ m =4.41 \ Km}$$ (2) Time taken to reach maximum height Time taken to reach maximum height is $${t=\frac{u}{g}=\frac{294}{9.8}=30 \ s}$$ (3) Velocity at the end…