Question

annie wasli memorial secondary school
sss2 science 1 first term physics test 2 (answer all questions) 2 hours
question 1
a stone is projected vertically upward with a speed of 30 ms⁻¹ from the top of a tower of height 50 m. neglecting air resistance, determine g = 10 ms⁻²
a. the maximum height reached from the ground.
b. the time of flight
c. the total distance
question 2
a. sketch velocity - time graph to illustrate each of the following:
i) uniform motion;
ii) uniform acceleration with an acceleration a m/s², a > 0;
iii) a body moving with zero acceleration from point a to point b and then decelerate to point c
b. with the an illustration how can we determine the
i) the acceleration
ii) deceleration
iii) total distance

Explanation:

Step1: Analyze the motion (upward)

The stone is projected upward with initial velocity \( u = 30\space m/s \), acceleration \( a=-g=- 10\space m/s^{2}\) (negative as it's opposite to motion). At maximum height, final velocity \( v = 0 \). Use the formula \( v^{2}=u^{2}+2as \) (where \( s \) is the height from the tower top, \( h_1 \)).

$$ 0=(30)^{2}+2(- 10)h_1 $$

$$ 20h_1 = 900 $$

$$ h_1=\frac{900}{20}=45\space m $$

Maximum height from ground: \( H=h_1 + 50=45 + 50 = 95\space m \)

Step2: Time to reach max height (upward)

Use \( v=u+at \). For upward motion to max height, \( v = 0 \), \( u = 30\), \( a=-10 \).

$$ 0 = 30-10t_1 $$

$$ t_1=\frac{30}{10}=3\space s $$

Step3: Time to fall from max height to ground

Height to fall \( H = 95\space m \), initial velocity for fall \( u'=0 \), \( a = g = 10\space m/s^{2}\). Use \( s=u't+\frac{1}{2}at^{2} \).

$$ 95=0+\frac{1}{2}(10)t_2^{2} $$

$$ 5t_2^{2}=95 $$

$$ t_2^{2}=19 $$

$$ t_2=\sqrt{19}\approx4.36\space s $$

Total time of flight \( t=t_1 + t_2=3 + 4.36 = 7.36\space s \) (approx, but let's check another way for total distance. Wait, total distance: upward distance \( 45\space m \), downward distance \( 95\space m \), so total distance \( D = 45+95 = 140\space m \). Wait, maybe re - check the first part for total distance. Wait, the problem has three parts: a) max height, b) time of flight, c) total distance.

Wait, maybe I messed up the total distance. Let's re - do:

For part a) Maximum height from ground:

Upward motion: \( v^{2}=u^{2}-2gh_1\) (taking upward as positive, \( g = 10\)). At max height, \( v = 0 \). So \( 0 = 30^{2}-2\times10\times h_1\Rightarrow h_1=\frac{900}{20}=45\space m \). So max height from ground is \( 50 + 45=95\space m \).

Part b) Time of flight: Time to go up (\( t_1 \)) and time to come down from max height to ground (\( t_2 \)).

Time to go up: \( v = u - gt_1\Rightarrow0 = 30-10t_1\Rightarrow t_1 = 3\space s \).

Time to come down: The stone falls from \( 95\space m \) with \( u = 0 \), \( s=95\space m \), \( a = 10\space m/s^{2}\). Use \( s=ut+\frac{1}{2}at^{2}\Rightarrow95 = 0+\frac{1}{2}\times10\times t_2^{2}\Rightarrow5t_2^{2}=95\Rightarrow t_2^{2}=19\Rightarrow t_2=\sqrt{19}\approx4.36\space s \). Total time \( t=t_1 + t_2=3 + 4.36 = 7.36\space s \) (approx). But maybe we can use the equation for displacement from the start point (tower top) to ground. The displacement \( s=- 50\space m \) (downward). Use \( s=ut+\frac{1}{2}at^{2}\Rightarrow-50 = 30t-5t^{2}\Rightarrow5t^{2}-30t - 50 = 0\Rightarrow t^{2}-6t - 10 = 0 \). Solve using quadratic formula \( t=\frac{6\pm\sqrt{36 + 40}}{2}=\frac{6\pm\sqrt{76}}{2}=\frac{6\pm2\sqrt{19}}{2}=3\pm\sqrt{19} \). Take positive root: \( t = 3+\sqrt{19}\approx3 + 4.36 = 7.36\space s \), same as before.

Part c) Total distance: Distance going up (\( 45\space m \)) and distance coming down (\( 95\space m \)), so total distance \( D = 45+95 = 140\space m \).

Answer:

a) Maximum height from ground: \(\boldsymbol{95\space m}\)

b) Time of flight: \(\boldsymbol{\approx7.36\space s}\) (or \(3 + \sqrt{19}\space s\))

c) Total distance: \(\boldsymbol{140\space m}\)