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3 years ago

An athlete performing a long jump leaves the ground at a 27.0 degree angle and lands 7.80m away. What was the takeoff speed?

Physics

1 answer:

Dima020 [189]3 years ago

3 0

Answer:

=9.72 m/s

Explanation:

From the Newton's laws of motion;

x=2(v²cos∅sin∅)/g

Using geometry we see that 2 cos∅sin∅ = sin 2∅

Therefore, x= (v²sin 2∅)g, where v is the take off speed x the range and ∅ the launch angle.

Making v the subject of the formula we obtain the following equation.

v=√{xg /(sin 2∅)}

x=7.80

∅=27.0

v=√{7.8×9.8/sin(27×2)}

v=√94.485

v=9.72 m/s

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||||| A 1.5 kg block and a 2.5 kg block are attached to opposite ends of a light rope. The rope hangs over a solid, frictionless

amid [387]

Answer

The lighter block will have the Positive Acceleration that is +2.45 m/s square.

A (1.5 Kg) = + 2.45 m/s square

Explanation:

To solve this there is a hard way to do this and there is an easy way to do this. The hard way is to solve Newton's second law for each block individually and then combine them and you get two equations with two unknowns. you try your best to solve the algebra without losing any sins but lets be honest it usually goes wrong.

So the easy way to do this the way to get the magnitude of the acceleration of the blocks. That is to say that i want to know the magnitude at which 2.5 Kg block accelerates or 1.5 Kg block accelerates when the the blocks were released.

Take the net external force that tries to make system go and divide it by total mass of the the system.

A (of System) = F(net external force) / m (total mass of system)

This is the quick way to know the magnitude of acceleration of the objects in the system. But this is only possible if the system is moved in same magnitude of acceleration that is 2.5 Kg block will move downward and 1.5 Kg block will move upward with the same magnitude. So here in this case we have friction-less pulley and the blocks will move with the same magnitude of acceleration.

To find the external forces

External forces are the forces which exerted on the objects in our system from the objects outside of our system. So one external force is the force of gravity. Both 2.5 Kg block and the force of gravity will be in downward direction.

Force of gravity on 2.5 Kg block

F = + (2.5 x 9.8) = 24.5

After releasing the rope the 2.5 Kg block will drive the system and accelerates in downward direction so It will be a positive force.

Force of gravity on 1.5 Kg block

F = - (1.5 x 9.8) = 14.7

The force of gravity on 1.5 Kg block will be negative because it will accelerate in upward or opposite direction of the force of gravity. Because the whole system is moving in one direction but the force of gravity on 1.5 Kg block is opposing the acceleration of the system.

Now divide the the Net external forces by total mass of the blocks that is

A (of System) = F(net external force) / m (total mass of system)

A = (+ 24.5 - 14.7) / 2.5 + 1.5

A = 9.8 / 4

A = 2.45 meter per second square

So the Acceleration of 2.5 Kg block will be Negative that is -2.45 m/s square. Since block is accelerating down and we usually treat down as negative.

A (1.5 Kg) = - 2.45 m/s square

So the Acceleration of 1.5 Kg block will be Positive that is +2.45 m/s square. Since block is accelerating up and we usually treat up as positive.

A (1.5 Kg) = + 2.45 m/s square

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