Question

a shell is fired at an angle of 15 degrees to the horizontal hits the ground 3 km away . at what angle it should be projected so that it can hit a target 8 km away .

a.30 °
b. 60 °
c. 45°
d. can't hit the target​

Answer 1

Answer:

Range R=u

2

Sin2θ/g=6000meter

so u

2

=120000m/s so u=346.4m/s

Note, we used Sin2θ=Sin30

0

=0.5

Now we know that the maximum value of range is given as R

max

=u

2

/g=120000/10=12000meter=12Km

So its possible to hit the target at 10Km distance because its less than maximum range.

Answered By

Answer 2

Given:

A shell is fired at an angle of 15 degrees to the horizontal hits the ground 3 km away .

To find:

At what angle it should be projected so that it can hit a target 8 km away ?

Calculation:

Let Initial velocity of projection be "u":

$\therefore \: \dfrac{ {u}^{2} \sin(2 \theta) }{g} = 3000$

$\implies \: \dfrac{ {u}^{2} \sin(2 \times {15}^{ \circ} ) }{g} = 3000$

$\implies \: \dfrac{ {u}^{2} \sin( {30}^{ \circ} ) }{g} = 3000$

$\implies \: \dfrac{ {u}^{2} \times \frac{1}{2} }{g} = 3000$

$\implies \: \dfrac{ {u}^{2} }{g} = 3000 \times 2$

$\implies \: \dfrac{ {u}^{2} }{g} = 6000 \: \: \: \: \: ......(1)$

Now , maximum range is obtained at 45°.

$\therefore \: R_{max} = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} \sin(2 \times {45}^{ \circ} ) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} \sin( {90}^{ \circ} ) }{g}$

$\implies \: R_{max} = \dfrac{ {u}^{2} }{g}$

$\implies \: R_{max} = 6000 \: m$

$\implies \: R_{max} = 6 \:k m$

So, the max range is 6 km.

Since target is located 8 km away, we can say that the projectile can't hit the target.