Name: Motion With Variable Acceleration
Uploaded: 2019-07-19
Duration: 10 min 54 s
Description: You may be aware of the SUVAT equation, but do you think this can be applicable for all types of motion? No, what are those motion? Watch this to find out.

Question

A Bomber flying upward at an angle of

53 °

with the vertical releases a bomb at an altitude of

80 m

. The bomb strikes the ground

20 s

after its release. Find: [Given

\sin 53 ° = 0.8; g = 10 {ms}^{- 2}

]

(i) The velocity of the bomber at the time of release of the bomb.

(ii) The maximum height attained by the bomb

(iii) The horizontal distance travelled by the bomb before its strikes the ground

(iv) The velocity (magnitude & direction) of the bomb just when it strikes the ground.

Accepted Answer

(i) Using the equation of motion, displacement along y-axis

$s_{y} = u_{y} + \frac{1}{2} a_{y} t^{2}$

$\Rightarrow - h = u \sin θ t - \frac{1}{2} g t^{2}$

$\Rightarrow - 800 = u \sin 37^{\circ} \times 20 - \frac{1}{2} \times 10 \times 20 \times 20$

$\Rightarrow - 80 = 2 u \sin 37^{\circ} - 200$

$\Rightarrow 2 u \sin 37^{\circ} = 120$

$\Rightarrow u \sin 37^{\circ} = 60 \Rightarrow u \times \frac{3}{5} = 60 \Rightarrow u = 100 {ms}^{- 1}$

(ii) Using equation of motion,

$v = u + g t \Rightarrow 0 = u \sin 37^{\circ} - g t \Rightarrow 0 = 100 \times \frac{3}{5} - 10 t \Rightarrow t = 6 \sec$

Height attained by ball is given by

$h = u \sin 37^{\circ} t - \frac{1}{2} g t^{2} \Rightarrow h = 100 \times \frac{3}{5} \times 6 - \frac{1}{2} \times 10 \times 6 \times 6 \Rightarrow h = 360 - 180 = 180 m$

So, maximum height will be

$H_{m a x} = 800 + 180 = 980 m$

(iii) Horizontal distance $=$ Horizontal velocity $\times$ Time of flight

$= 100 \cos 37 ° \times 20 = 100 \times \frac{4}{5} \times 20 = 1600 m$

(iv) Initial velocity can be written as

$\vec{u} = u \cos θ \hat{i} + u \sin θ \hat{j}$

$\Rightarrow \vec{u} = 100 \times \frac{4}{5} \hat{i} + 100 \times \frac{3}{5} \hat{j}$

$\Rightarrow \vec{u} = 80 \hat{i} + 60 \hat{j}$

Final velocity is given by

$\vec{v} = 60 - 10 \times t$

As $t = 20 \sec$

$\vec{v} = 60 - 10 \times 20 = 60 - 200, \vec{v} = - 140 {ms}^{- 1}$

${\vec{v}}_{x} = 80 {ms}^{- 1}$

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