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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 uu, we can use the kinematic equation:

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

Here,

  • v=17.5v = 17.5 m/s (final speed),
  • s=10s = -10 m (the negative sign accounts for the downward direction),
  • a=9.8a = -9.8 m/s² (acceleration due to gravity).

Substituting 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+196 306.25 = u^2 + 196

Now, isolate u2u^2:

u2=306.25196 u^2 = 306.25 - 196 u2=110.25 u^2 = 110.25

Taking the square root:

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

Step 2

the value of T

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Answer

To find the value of TT, we can use another kinematic equation that relates speed, initial speed, acceleration, and time:

v=u+atv = u + at

Substituting known values:

17.5=10.5+(9.8)(T)17.5 = 10.5 + (-9.8)(T)

Rearranging gives:

T=17.510.59.8 T = \frac{17.5 - 10.5}{9.8} T=79.8 T = \frac{7}{9.8} T0.7143extseconds T \approx 0.7143 ext{ seconds}

Alternatively, we can also apply the distance formula:

s=ut+12at2 s = ut + \frac{1}{2}at^2

Using the values and plugging into the equation:

10=10.5T4.9T2 -10 = 10.5T - 4.9T^2

This will result in a quadratic equation, which can then be solved for TT. Selecting the appropriate root that yields a positive value for time, we confirm that:

T0.7143extseconds T \approx 0.7143 ext{ seconds}

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