Aprojectile is launched straight up at 141 m/s . how fast is it moving at the top of its trajectory? suppose it is launched upward at 45 degree above the horizontal plane. how fast is it moving at the top of its curved trajectory?

The velocity of projectile has 2 components, horizontal component vcosθ and vertical component vsinθ, where v is the velocity of projection and θ is the angle between +ve X-axis and projectile motion.

In case 1, θ = 90⁰

    So horizontal component is vcos90 = 0

         Vertical component at maximum height = 0

So velocity at maximum height = 0 m/s

In case 2, θ = 45⁰

    So horizontal component is 141cos45 = 100m/s

         Vertical component at maximum height = 0

So velocity at maximum height = 100 m/s

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

in simple terms, the principle states that the buoyancy force on an object is equal to the weight of the fluid displaced by the object, or the density of the fluid multiplied by the submerged volume times the gravitational acceleration, g. this is also known as upthrust.