smc-4273-50-s=ut+1/2at^2

A hot-air balloon is travelling at a constant upwards velocity of 15 ms\(^{-1}\).

A passenger on the hot-air balloon decides to time how long it takes a pen to hit the ground when dropped from a height of 50 m.

Ignoring air resistance, determine how long it will take the pen to hit the ground. (4 marks)

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\(5.07\ \text{s}\)

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\(s= 50, \ a=9.8, \ u=-15\)

Downwards \(\Rightarrow\) positive direction

\(s\)	\(=ut+\dfrac{1}{2}at^2\)
\(50\)	\(=-15t + \dfrac{1}{2} \times 9.8 \times t^2\)
\(0\)	\(=4.9t^2-15t-50\)
\(t\)	\(=\dfrac{15\pm \sqrt{225-4 \times 9.8 \times -50}}{2 \times 4.9}\)
	\(=5.07, -2.01\)

Time for the pen to fall = 5.07 seconds \( (t \gt 0) \).

A cricket ball is hit vertically upwards from ground level, it gains 70 metres vertically and then falls back to the ground.

At what time(s) will the ball be 40 metres above the ground? Ignore any air resistance, giving your answer in seconds, correct to two decimal places. (4 marks)

--- 7 WORK AREA LINES (style=lined) ---

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\(6.25\ \text{s}\ \ \text{and}\ \ 1.31\ \text{s}\)

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\(\text{At 70 m}\ \Rightarrow \ v=0\ \text{ms}^{-1}:\)

\(v^2\)	\(=u^2+2as\)
\(0^2\)	\(=u^2+2 \times -9.8 \times 70\)
\(u^2\)	\(=1372\)
\(u\)	\(=37.04\ \text{ms}^{-1}\)

\(\text{Find}\ t\ \text{when}\ \ s=40:\)

\(s\)	\(=ut+\dfrac{1}{2}at^2\)
\(40\)	\(=37.04t+\dfrac{1}{2} \times -9.8 \times t^2\)
\(4.9t^2\)	\(-37.04t+40=0\)
\(t\)	\(=\dfrac{37.04 \pm \sqrt{(-37.04)^2-4 \times 4.9 \times 40}}{2 \times 4.9}\)
	\(=6.25\ \text{s}\ \ \text{and}\ \ 1.31\ \text{s}\)

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