The speed which a ball must be thrown to reach 8m.

In summary, an object with a mass of 0.2 kg is thrown vertically upwards from ground level and reaches a height of 8.0 m. To calculate the speed at which it must be thrown, we can use the equation Vf2 – Vo2 = 2gH, where Vf is the final velocity (in this case, 0), Vo is the initial velocity, g is the acceleration due to gravity (9.8 m/s2), and H is the height (8.0 m). We can rearrange the equation to solve for Vo, and with the correct units, we find that the initial velocity must be 12.52 m/s.
  • #1
Meezus
57
0

Homework Statement



An object of mass 0.2 kg is thrown vertically upwards from ground level and reaches a height of 8.0 m. Calculate: the speed with which it must be thrown.

Homework Equations



Vf2 – Vo2 = 2gH

The Attempt at a Solution



Vf2 – Vo2 = 2gH
Vf = Final Velocity = 0 (Ball at maximum height)
Vo = Initial Velocity.
g = 9.8
H =8m
0 - Vo2 = 2 x -9.8 x 8 = -156.8

This is my attempt so far but I'm getting a negative answer, can anyone show me where I'm going wrong?
 
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  • #2
Suggesting there is no air resistance ( no loss of energy ):

½*m*V02 = m*g*H →

V02 = 2*g*H

or you could say:

V02 – Vf2 = 2*g*H ( Vf = 0 ).
 
  • #3
Hesch said:
Suggesting there is no air resistance ( no loss of energy ):

½*m*V02 = m*g*H →

V02 = 2*g*H

or you could say:

V02 – Vf2 = 2*g*H ( Vf = 0 ).

This is what I'm already doing but getting a negative number? Should I just be times by 9.8 not -9.8?
 
  • #4
No, I have switched the order of velocities, V0 and Vf .
This will change the sign of the result.
 
  • #5
And you should use g = +9.8 m/s2.
Otherwise the potential energy will be negative.
 
  • #6
Hesch said:
And you should use g = +9.8.
Otherwise the potential energy will be negative.

so V02 – Vf2 = 2*g*H ( Vf = 0 ).
0 - Vf2 = 2 x 9.8 x 8 = 156.8.

is this correct?
 
  • #7
No, your first line is correct, but the second is not.

V02 is not equal to 0.
 
  • #8
Hesch said:
No, your first line is correct, but the second is not.

V02 is not equal to 0.

V02 = 2*g*H

so V02= 2 x 9.8 x 8 = 156.8
 
  • #9
Yes.

Remember units!
 
  • #10
Hesch said:
Yes.

Remember units!

156.8 m/s correct?
and thank you
 
  • #11
No, try to insert units in this calculation:

2 x 9.8 x 8 = 156.8

Will the unit of the result be m/s? Or will it be m2/s2?
Why?
 
  • #12
V02 = 2 × 9.8 [m/s2] × 8 [m] = 156.8 [m2/s2] →
V0 = (156.8 [m2/s2]) = 12.52 [m/s]

Always check the units. Then you can see if something is wrong in your calculation.
Here the square root was missing.
 
  • #13
Hesch said:
V02 = 2 × 9.8 [m/s2] × 8 [m] = 156.8 [m2/s2] →
V0 = (156.8 [m2/s2]) = 12.52 [m/s]

Always check the units. Then you can see if something is wrong in your calculation.
Here the square root was missing.
Thank you for all your help :) it is greatly appreciated
 

What is the formula for calculating the speed needed to reach 8m?

The formula for calculating the speed needed to reach 8m is v = √(2gh), where v is the speed in meters per second, g is the acceleration due to gravity (9.8 m/s²), and h is the height in meters.

What factors affect the speed needed to reach 8m?

The factors that affect the speed needed to reach 8m include the height from which the ball is thrown, the acceleration due to gravity, and any air resistance or other external forces acting on the ball.

What is the minimum speed required to reach 8m?

The minimum speed required to reach 8m is approximately 8.86 m/s, assuming the ball is thrown from ground level and there is no air resistance or other external forces acting on the ball.

What is the maximum speed that a ball can be thrown to reach 8m?

The maximum speed that a ball can be thrown to reach 8m is infinite, as long as the height from which it is thrown is also infinite. However, in a real-world scenario, factors such as air resistance and the physical limitations of the thrower will limit the maximum possible speed.

What are some real-world applications of calculating the speed needed to reach 8m?

Calculating the speed needed to reach 8m can be useful in a variety of sports and activities, such as throwing a javelin or discus, hitting a golf ball, or performing a long jump. It can also be applied in engineering and physics, for example in designing roller coasters or calculating the trajectory of a projectile.

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