Question

a ball is thrown upwards at an angle of 60 degree to the horizontal.it falls on the ground at distance of 90m.if the ball is thrown with same initial velocity and 30 degree,it will fall on ground at a distance of
a}120 m
b]90
c}30
d}60

Answer 1

$90 = \frac{u^2*sin2(60^{o}) }{g} \\ g=10 m/s^{2} \\ u^{2} = \frac{1800}{\sqrt{3} }$

Putting the value into the new equation
$R = \frac{\frac{1800}{\sqrt{3}}*sin 2(30^{o}) } {g} \\ R =90$

Answer 2

Option b -90 m is the distance in which it fall on ground.

Given:

60°  a ball is “thrown upwards”.

Distance on which the ball falls = 90 m

The ball is thrown with the same initial velocity = 30°

To find:

The distance in which in falls on the ground = ?

Solution:  

The formula for distance covered by the ball is given as  $x=\frac{v^{2} \sin 2 \theta}{g}$ where v is the velocity at which the ball is thrown, g is the gravitational force of attraction towards earth, which is considered as $10m/sec^2$ .  

Given that ball is thrown at 60°  angle to the horizontal,

Given x = 90m, θ= 60°  

Substituting the values of the above in $x=\frac{v^{2} \sin 2 \theta}{g}$,

$90=\frac{v^{2} \sin ^{2} 60}{10} \Rightarrow 900=\left(v^{2} \sin 120\right)=\left(v^{2} \times \frac{\sqrt{3}}{2}\right)=\sqrt{3} v^{2}$  (As sin120 is same as sin60 –sine function is positive in first and second quadrant)

$v^{2}=\frac{1800}{\sqrt{3}} -------(1)$

Now, the ball is thrown with same velocity and with 30 degree angle, we need to find the distance it covers on the ground. As the velocity is same, we can use the same velocity to find the distance covered when the ball is thrown at an angle of 30 degree as follows:

$x=\frac{v^{2} \sin 2 \theta}{g}$

$x=\left(\left(\frac{1800}{\sqrt{3}}\right) \times \sin ^{2} \frac{(30)}{10}\right)$

(from (1)$v^2$ value is taken)

$x=\frac{\frac{1800}{\sqrt{3}} \times \frac{\sqrt{3}}{2}}{10}$

$x=90 \mathrm{m}$

Therefore, the ball will fall at a distance of 90m.