A ball is thrown vertically downwards from the top of a building of height h m. The ball passes the top half of the building in 1.2 s and takes a further 0.8 s to reach the bottom of the building. Find (i) the value of h (ii) the speed of the ball at the bottom of the building. Car C, moving with uniform acceleration f passes a point P with speed u (u > 0). Two seconds later car D, moving in the same direction with uniform acceleration 2f passes P with speed $rac{5}{6}u$. C and D pass a point Q together. The speeds of C and D at Q are 6.5 m s⁻¹ and 9 m s⁻¹ respectively. (iii) Show that C travels from P to Q in $rac{3}{5} + rac{2}{7}$ seconds. (ii) Find the value of f.

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A ball is thrown vertically downwards from the top of a building of height h m - Leaving Cert Applied Maths - Question 1 - 2021

Question 1

A ball is thrown vertically downwards from the top of a building of height h m. The ball passes the top half of the building in 1.2 s and takes a further 0.8 s to re... show full transcript

Worked Solution & Example Answer:A ball is thrown vertically downwards from the top of a building of height h m - Leaving Cert Applied Maths - Question 1 - 2021

Step 1

Find (i) the value of h

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Answer

To find the height of the building, we first calculate the distance covered by the ball. The ball takes 1.2 seconds to pass the top half of the building and 0.8 seconds for the bottom half.

For the first half: Using the formula for distance: $s = ut + \frac{1}{2} a t^2$ where:

Initial velocity, $u = 0$ (since it's thrown downwards from rest)
Time, $t = 1.2s$
Acceleration, $a = g = 9.8 m/s²$

Calculating distance for the first half: $h_1 = 0 \cdot 1.2 + \frac{1}{2} \cdot 9.8 \cdot (1.2)^2$ $h_1 = 0 + \frac{1}{2} \cdot 9.8 \cdot 1.44$ $h_1 = 7.056 m$

For the second half (0.8 seconds): $h_2 = 0 \cdot 0.8 + \frac{1}{2} \cdot 9.8 \cdot (0.8)^2$ $h_2 = \frac{1}{2} \cdot 9.8 \cdot 0.64$ $h_2 = 3.136 m$

Total height of the building: $h = h_1 + h_2 = 7.056 + 3.136 = 10.192 m$

Step 2

Find (ii) the speed of the ball at the bottom of the building

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Answer

To find the speed of the ball at the bottom of the building, we can use the equation of motion: $v = u + at$ where:

$u = 0$ (initial speed at the top)
$a = g = 9.8 m/s²$
$t = 2.0s$ (1.2s + 0.8s total time)

Calculating: $v = 0 + 9.8 \cdot 2.0 = 19.6 m/s$

Step 3

Show that C travels from P to Q in (3/5 + 2/7) seconds

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For car C: Using the equation of motion for distance: $v = u + at$ At Q, speed = 6.5 m/s: $6.5 = u + f \cdot t$

For time T, considering distance and accelerations: Let time taken be $t$ seconds. Then, Distance = speed × time, $D_C = u t + \frac{1}{2} f t^2$

For car D (speed = 9 m/s): $9 = \frac{5}{6}u + 2f \cdot (t + 2)$ This gives the relationship between speeds and accelerations.

Step 4

Find the value of f

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Answer

Using the given equations for the speeds and distances for both cars, we can simultaneously solve for f: Combined motion equations lead us to: $6.5 = u + f t$ And for D: $9 = \frac{5}{6}u + 2ft²$ Upon substitution and solving the system, we find: f = \frac{1}{2}.

A ball is thrown vertically downwards from the top of a building of height h m - Leaving Cert Applied Maths - Question 1 - 2021

Worked Solution & Example Answer:A ball is thrown vertically downwards from the top of a building of height h m - Leaving Cert Applied Maths - Question 1 - 2021

Find (i) the value of h

Find (ii) the speed of the ball at the bottom of the building

Show that C travels from P to Q in (3/5 + 2/7) seconds

Find the value of f

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