Question

A missile is fired from the ground level rises x meters vertically upwards in t sec, where \(x=100t-\frac{25}{2}t^2\). the maximum height reached is

• A. 100m
• B. 300m
• C. 125m
• D. 200m

Answer 1

The Correct Option is D

Given:
Displacement function:
\[ x = 100t - \frac{25}{2}t^2 \] 
Step 1: Velocity as a function of time 
\[ \frac{dx}{dt} = 100 - 25t \]
Initial velocity:
At \(t = 0\):
\[ v = 100 - 25 \cdot 0 = \boxed{100\,\text{m/s}} \]
Step 2: Maximum height
At maximum height, velocity \(v = 0\):
\[ 100 - 25t = 0 \Rightarrow t = 4\,\text{s} \]
Substitute \(t = 4\) into displacement equation:
\[ x = 100 \cdot 4 - \frac{25}{2} \cdot 16 = 400 - 200 = \boxed{200\,\text{m}} \]
Step 3: Time to return to ground
Displacement becomes zero when missile returns to ground:
\[ x = 100t - \frac{25}{2}t^2 = 0 \Rightarrow t(100 - \frac{25}{2}t) = 0 \Rightarrow t = 0 \text{ or } t = 8 \]
Step 4: Velocity when missile hits the ground
At \(t = 8\):
\[ \frac{dx}{dt} = 100 - 25 \cdot 8 = -100\,\text{m/s} \Rightarrow \text{Same magnitude as initial velocity, but negative (downward).} \]
Step 5: Average speed during the entire motion
Total distance traveled = \(200\,\text{m (up)} + 200\,\text{m (down)} = 400\,\text{m}\)
Total time = \(8\,\text{s}\)
\[ \text{Average speed} = \frac{400}{8} = \boxed{50\,\text{m/s}} \]
But if the question asked for displacement magnitude from launch to hit:
\[ \text{Net displacement} = 0 \Rightarrow \text{Distance from highest point (200 m) to ground: } \boxed{200\,\text{m}} \]
If instead the question asked for displacement at \(t = 5\):
\[ x = 100 \cdot 5 - \frac{25}{2} \cdot 25 = 500 - 312.5 = \boxed{187.5\,\text{m}} \]
Final Answer Based on Given Calculation:
\[ \boxed{200\,\text{m}} \text{ is the correct maximum height} \]