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At time t = 0, a particle is projected vertically upwards with speed u m s⁻¹ from a point 10 m above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2008 - Paper 1

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At time t = 0, a particle is projected vertically upwards with speed u m s⁻¹ from a point 10 m above the ground. At time T seconds, the particle hits the ground with... show full transcript

Worked Solution & Example Answer:At time t = 0, a particle is projected vertically upwards with speed u m s⁻¹ from a point 10 m above the ground - Edexcel - A-Level Maths Mechanics - Question 2 - 2008 - Paper 1

Step 1

the value of u

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Answer

To find the value of u, we can use the equation of motion:

v2=u2+2asv^2 = u^2 + 2as

Where:

  • Final velocity, v = 17.5 m/s (downwards, hence taken as negative)
  • Initial velocity, u = ?
  • Acceleration, a = -9.8 m/s² (due to gravity)
  • Displacement, s = -10 m (the particle is falling to the ground)

Substituting these values into the equation:

(17.5)2=u2+2(9.8)(10)(-17.5)^2 = u^2 + 2(-9.8)(-10)

Calculating the left side:

306.25=u2+196306.25 = u^2 + 196

Now, rearranging the equation:

u2=306.25196u^2 = 306.25 - 196

u2=110.25u^2 = 110.25

Taking the square root gives:

u=extsqrt(110.25)=10.5extm/su = ext{sqrt}(110.25) = 10.5 ext{ m/s}

Step 2

the value of T

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Answer

For part (b), we can again use the equation of motion:

v=u+atv = u + at

Here:

  • Final velocity, v = -17.5 m/s
  • Initial velocity, u = 10.5 m/s
  • Acceleration, a = -9.8 m/s²
  • Time, T = ?

Substituting these values into the equation:

17.5=10.59.8T-17.5 = 10.5 - 9.8T

Rearranging gives:

17.510.5=9.8T-17.5 - 10.5 = -9.8T

28=9.8T-28 = -9.8T

Dividing both sides by -9.8:

T = rac{28}{9.8}

Calculating gives:

T ext{ approximately equals } 2.857 ext{ seconds (or } rac{20}{7} ext{ seconds)}

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