Question

3. It is observed that for two different angles of projection the horizontal
range is same. If the maximum vertical distantces covered in the two
cases are respectively 36 and 25 metre, then calculate the horizontal
range.
​

Answer 1

Answer: 120m

Explanation:

Range=4√H1×H2

=4√36×25

=4×6×5

=120m

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Answer 2

The horizontal range is 120 m.

Given,

For two different angles of projection the horizontal range is same.

Maximum vertical distance(maximum height) covered in the two

cases are respectively 36 and 25 metre.

To find,

the horizontal range.

Solution:

• The motion here referred to is a projectile motion.
• When a particle is thrown with velocity u in a projectile motion, the maximum height(maximum vertical height), H attained by the particle is given as:
• $H=\frac{u^{2} sin^{2} (theta)}{2g}$.
• where, theta-angle with the horizontal and g-acceleration due to gravity.
• g=10 $m/s^{2}$.
• When a particle is thrown with velocity u in a projectile motion, the horizontal range, R covered by the particle is given as:
• $R=\frac{u^{2}sin 2(theta) }{g} =\frac{u^{2}2sin(theta)cos(theta) }{g}$.
• For angles of projection θ and (90-θ), the horizontal ranges are same.

Thus let us assume that particle 1 is thrown with a velocity u that makes an angle θ with the horizontal and particle 2 is also thrown with a velocity u that makes an angle (90-θ) with the horizontal.

For particle 1 thrown with angle θ,

$H=\frac{u^{2} sin^{2} (theta)}{2g}\\36=\frac{u^{2} sin^{2} (theta)}{2*10}\\36*20=u^{2} sin^{2} (theta)\\720=u^{2} sin^{2} (theta)\\usin(theta)=\sqrt{720} .$

For particle 2 thrown with angle (90-θ),

$H=\frac{u^{2} sin^{2} (theta)}{2g}\\25=\frac{u^{2} sin^{2} (90-theta)}{2*10}\\25*20=u^{2} sin^{2} (90-theta)\\5000=u^{2} sin^{2} (90-theta)\\ucos(theta)=\sqrt{500} .$

The range will be:

$R=\frac{u^{2}sin 2(theta) }{g} =\frac{u^{2}2sin(theta)cos(theta) }{g}\\R=\frac{u^{2}2sin(theta)cos(theta) }{g}\\R=\frac{2(usin(theta))(ucos(theta))}{g} \\R=\frac{2*\sqrt{720}\sqrt{500} }{10} \\R=\frac{600*2}{10} \\R=60*2 m\\R=120 m.$

Hence, the range is 120 m.

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