Question

please help :)
A small ball is projected with speed Ums from a point O at the top of a vertical cliff.
The point is 25 m vertically above the point N which is on horizontal ground.
The ball is projected at an angle of 45° above the horizontal.
The ball hits the ground at a point A, where AN= 100 m, as shown in Figure 2.
The motion of the ball is modelled as that of a particle moving freely under gravity.
Using this initial model,
(a) show that U28

(b) find the greatest height of the ball above the horizontal ground NA.

In a refinenment to the model of the motion of the ball from O to A, the effect of air
resistance is included.

This refined model is used to find a new value of U.
(c) How would this new value of U compare with 28, the value given in part (a)?

(d) State one further refinement to the model that would make the model more realistic.
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Answer 1

Answer:

please help :)

A small ball is projected with speed Ums from a point O at the top of a vertical cliff.

The point is 25 m vertically above the point N which is on horizontal ground.

The ball is projected at an angle of 45° above the horizontal.

The ball hits the ground at a point A, where AN= 100 m, as shown in Figure 2.

The motion of the ball is modelled as that of a particle moving freely under gravity.

Using this initial model,

(a) show that U28

(b) find the greatest height of the ball above the horizontal ground NA.

In a refinenment to the model of the motion of the ball from O to A, the effect of air

resistance is included.

This refined model is used to find a new value of U.

(c) How would this new value of U compare with 28, the value given in part (a)?

(d) State one further refinement to the model that would make the model more realistic.

Answer 2

Answer:

A)$U=28$ as required proved

B) Height is $45 m$

C) The new value of U is larger than 28 to compensate for the wind resistance

D) Account the spining of the ball

Step-by-step explanation:

a) $(\rightarrow) motion 100=U \cos 45^{\circ} x t ....... (I)$

$\text { (+) tve } \begin{aligned}v &=-25 \mathrm{~m} \\v &=u \sin 45 \\v &=?\\a &=-9.8 \mathrm{~ms}^{-2} \\t &=\end{aligned}$

$\begin{aligned}&s=u t+\frac{1}{2} a t^{2} \\&-25=4 \sin 45 \times t+\frac{1}{2}(-9 \cdot 8) t^{2}..........(2)\end{aligned}$

$\begin{aligned}&\text { (1) gives } u=\frac{100}{\cos 45 \times t}\\&-25=\frac{100}{t} \times \frac{\sin 45}{\cos 45} t-4 \cdot 9 t^{2}\\&-25=100-4 \cdot 9 t^{2} \quad \mid \begin{array}{l|}\tan 45=1 \\\frac{\sin 45}{\cos 45}=\tan 45\\\end{array}\\&t=5.050762723,-5.05076272\\&\text { in } equation 1 . U=\frac{100}{5.050762723 \times \cos 45}\\&U=28 \text { (as required) }\end{aligned}$

b)

$\begin{aligned}} \\s={ ? }\\\&u=28 \sin 45 \mathrm{~ms}^{-1}\\\\&v=0 \mathrm{~ms}^{-1}\\&a=-9.8 \mathrm{~ms}^{-2}\\&t=2 \text {. }\\&v^{2}=u^{2}+2 a s\\&s=\frac{v^{2}-u^{2}}{2 a}=\frac{0^{2}-(28 \sin 45)^{2}}{2 x-9.8}\\&s=20 \mathrm{~m}\\&\text { Height above ground }=20+25=45 \mathrm{~m}\end{aligned}$

c) To reach paint $A$ the new value of $U$ would have to be larger than 28 to compensate for wind resistance.

d) Account for effects of wind or Use more accurate value for 9 or Account for spin of the ball