Question

A projectile is projected with certain speed at an angle of 45\(^\circ\) with horizontal as shown. At t = 2s, projectile is at maximum height and at t = 3s, it just touches a wall at a height H above horizontal. Find H in meters:
(Assume g = 10 m/s\(^2\))

• A. 20 m
• B. 10 m
• C. 15 m
• D. 25 m

Hint:

For projectile motion problems, breaking the motion into horizontal and vertical components is key. The time to reach maximum height is a crucial piece of information as it directly gives the initial vertical velocity (\(u_y\)). Analyzing motion relative to the peak can sometimes simplify calculations for the second half of the trajectory.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A projectile reaches its maximum height at t=2s. We need to find its vertical height (H) at a later time, t=3s.
Step 2: Key Formula or Approach:
1. The time to reach the maximum height for a projectile is given by \(t_{top} = u_y / g\), where \(u_y\) is the initial vertical component of velocity.
2. The vertical position (height) of a projectile at any time t is given by the kinematic equation: \(y(t) = u_y t - \frac{1}{2} g t^2\).
Step 3: Detailed Explanation:
Find the initial vertical velocity (\(u_y\)):
We are given that the time to reach maximum height is \(t_{top} = 2\) s.
Using the formula \(t_{top} = u_y / g\):
\[ 2 = \frac{u_y}{10} \] \[ u_y = 2 \times 10 = 20 \text{ m/s} \] Find the height H at t = 3 s:
Now we use the equation for vertical position to find the height H at t = 3 s.
\[ H = y(3) = u_y(3) - \frac{1}{2} g (3)^2 \] Substitute the values of \(u_y\) and g:
\[ H = (20)(3) - \frac{1}{2} (10) (9) \] \[ H = 60 - 5 \times 9 \] \[ H = 60 - 45 \] \[ H = 15 \text{ m} \] Alternative Method (motion from the peak):
At t=2s, the projectile is at its peak. The vertical velocity at the peak is zero. We can analyze the vertical motion for the next 1 second (from t=2s to t=3s).
Maximum height reached, \(H_{max} = u_y t_{top} - \frac{1}{2} g t_{top}^2 = (20)(2) - \frac{1}{2}(10)(2^2) = 40 - 20 = 20\) m.
The vertical distance fallen from the peak in \(\Delta t = 3 - 2 = 1\) s is:
\[ \Delta y = (0)(\Delta t) + \frac{1}{2} g (\Delta t)^2 = \frac{1}{2}(10)(1^2) = 5 \text{ m} \] The height H at t=3s is the maximum height minus the distance fallen.
\[ H = H_{max} - \Delta y = 20 - 5 = 15 \text{ m} \] Step 4: Final Answer:
The height H is 15 m.