Question

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an initial angle of 51.0 ◦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 . a) Neglecting air resistance, find the shell’s horizontal range. Answer in units of m.

Answer 1

Answer:

The horizontal range will be $2.55\times 10^5m$

Explanation:

We have given initial speed of the shell u = $1.6\times 10^3m/sec$

Angle of projection = 51°

Acceleration due to gravity $g=9.8m/sec^2$

We have to find maximum range

Horizontal range in projectile motion is given by

$R=\frac{u^2sin2\Theta }{g}=\frac{(1.60\times 10^3)^2sin(2\times 51^{\circ})}{9.81}=2.55\times 10^5m$

So the horizontal range will be $2.55\times 10^5m$

Answer 2

We have that for the Question it can be said that Neglecting air resistance, the shell’s horizontal range.

R=2.55*10^5m

From the question we are told

A shell is fired from the ground with an initial speed of 1.60 × 103 m/s (approximately five times the speed of sound) at an initial angle of 51.0 ◦ to the horizontal. The acceleration of gravity is 9.81 m/s 2 . a) Neglecting air resistance, find the shell’s horizontal range. Answer in units of m.

Generally the equation for the Horizontal Range  is mathematically given as

$R=\frac{u^\sin2\theta}{g}\\\\Therefore\\\\R=\frac{(1.60 * 10^3)^2sin*51}{9.81}$

R=2.55*10^5m

Therefore

Neglecting air resistance, the shell’s horizontal range.

R=2.55*10^5m

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