At time $t = 0$, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2 m s$^{-1}$ from a point 50 m above the horizontal ground. The ball B is projected vertically upwards from the ground with speed 20 m s$^{-1}$. At time $t = T$ seconds, the two balls are at the same vertical height, $h$ metres, above the ground. The balls are modelled as particles moving freely under gravity. Find a) the value of $T$, b) the value of $h$.

A-Level Edexcel Maths Mechanics Constant Acceleration - 1D: At time $t = 0$, two balls A and

At time $t = 0$, two balls A and B are projected vertically upwards - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 2

Question 4

At time $t = 0$, two balls A and B are projected vertically upwards. The ball A is projected vertically upwards with speed 2 m s$^{-1}$ from a point 50 m above the h... show full transcript

Worked Solution & Example Answer:At time $t = 0$, two balls A and B are projected vertically upwards - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 2

Step 1

the value of $T$

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Answer

To solve for $T$ , we start with the equations of motion for both balls, where the displacement $s = ut + \frac{1}{2} a t^2$ .

For Ball A:

Initial velocity ( $u_A$ ) = 2 m s $^{-1}$ ,
Initial height = 50 m,
Acceleration due to gravity ( $a$ ) = -9.8 m s $^{-2}$ (acting downwards).

The equation for ball A becomes:

s_A &= u_A t + \frac{1}{2} a t^2 \ s_A &= 50 + 2T - 4.9T^2 \ \text{(1)} \end{align*}$$ For Ball B: - Initial velocity ($u_B$) = 20 m s$^{-1}$, - Initial height = 0 m. The equation for ball B becomes: $$egin{align*} s_B &= u_B t + \frac{1}{2} a t^2 \ s_B &= 20T - 4.9T^2 \ \text{(2)} \end{align*}$$ At time $T$, both balls are at the same height, thus: $$s_A = s_B$$ $$50 + 2T - 4.9T^2 = 20T - 4.9T^2$$ The $-4.9T^2$ cancels out: $$50 + 2T = 20T$$ Solving for $T$: $$50 = 20T - 2T$$ $$50 = 18T$$ $$T = \frac{50}{18} \approx 2.777...$$ Thus, rounding to two decimal places, $T \approx 2.78$ s.

Step 2

the value of $h$

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Answer

To find the height $h$ , substitute $T \approx 2.78$ into either of the equations of motion. We'll use the equation for ball B:

$h = s_B = 20T - 4.9T^2$ Substituting $T$ : $h = 20(2.78) - 4.9(2.78^2)$ Calculating:

First calculate $20(2.78) = 55.6$ .

Next, calculate $2.78^2 \approx 7.7284$ and $4.9(7.7284) \approx 37.87276$ .

So, $h = 55.6 - 37.87276 \approx 17.72724$ Rounding to 2 significant figures, we find: $h \approx 17.7 \text{ m}.$

At time $t = 0$, two balls A and B are projected vertically upwards - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 2

Worked Solution & Example Answer:At time $t = 0$, two balls A and B are projected vertically upwards - Edexcel - A-Level Maths Mechanics - Question 4 - 2013 - Paper 2

the value of $T$

the value of $h$

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