Question

A bullet is fired from a gun at the speed of 280 ms^-1 in the direction 30° above the horizontal. The maximum height attained by the bullet is
(g = 9.8 ms^-2, sin 30° = 0.5)

• A. 3000 m
• B. 2800 m
• C. 2000 m
• D. 1000 m

Answer 1

The Correct Option is D

Given: 
\(u = 280\ ms^{-1}\)
\(\theta = 30\degree\)
\(g-9.8 \ ms^{-2}\)

The maximum height attained by the bullet,
\(H_{max}=\frac{u^2sin^2\theta}{2g}\)

\(H_{max}=\frac{(280)^2\times(sin30^{\circ})^2}{2(9.8)}\)

\(H_{max}=1000\,m\)

So, the correct option is (D): \(1000\ m\)

Answer 2

To find the maximum height attained by the bullet, we can use the fact that at the highest point of its trajectory, the vertical component of the bullet's velocity will be zero. This is because, at this point, the bullet will have reached its maximum height and will momentarily stop before starting to fall back down.
First, we need to find the initial vertical and horizontal components of the velocity of the bullet. We can use trigonometry to do this:
Initial vertical velocity, \(v_y = 280 sin 30° = 140 \ m/s\)
Initial horizontal velocity, \(vx = 280 cos 30° = 242.4\ m/s\)
Next, we can use the fact that the maximum height attained by the bullet is equal to the vertical displacement of the bullet from its initial position. We can use the kinematic equation:
\(y = y_0 + v_yt - \frac{1}{2gt^2}\)
Where \(y_0\) is the initial vertical position (which we can take to be zero), t is the time taken to reach maximum height, and g is the acceleration due to gravity.
At the maximum height, \(v_y = 0,\) so we can rearrange the equation to solve for t:
\(t =\)\(\frac{v_y}{g}\)
Substituting in the values we have:
t = \(\frac{140}{9.8}\) = 14.29 s (to two decimal places)
Now we can use this value of t to find the maximum height:
\(y = y_0 + v_yt - \frac 12gt^2 y = 0 + 140(14.29) - \frac 12(9.8)(14.29)^2 y = 999.86 m\) (to two decimal places)

Therefore, the maximum height attained by the bullet is approximately 999.86 meters, equivalent to \(1000\ m\).