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asambeis [7]
3 years ago
5

If a ball is thrown up at an initial speed of 40. m/s, how many seconds does it take to reach the top of its path?

Physics
1 answer:
natka813 [3]3 years ago
5 0

Answer:

Time, t = 4.08 secs

Explanation:

<u>Given the following data;</u>

Initial velocity, U = 40m/s

To find the time, we would use the first equation of motion;

V = U + at

Where;

  • V is the final velocity.
  • U is the initial velocity.
  • a is the acceleration.
  • t is the time measured in seconds.

<em>Making time, t the subject of formula, we have;</em>

t = \frac{V - U}{a}

We know that acceleration due to gravity, g is 9.8m/s².

a = g = - 9.8m/s² because the ball is thrown in the opposite direction.  

Also, the final velocity is equal to zero (0) because the ball reached its maximum height.

<em>Substituting into the equation, we have;</em>

t = \frac{0 - 40}{-9.8}

t = \frac{-40}{-9.8}

Time, t = 4.08 secs

<em>Therefore, it will take the ball 4.08 seconds to reach the top. </em>

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m = 1.26*10²⁵ kg.

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Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

We know that the stone is thrown upward with an initial velocity vo = 15 m/s.

At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.

With this value for  rp, we can solve (2) for mp, as follows:

mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²

mp = 1.26*10²⁵ kg.

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3 years ago
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