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: Key idea used

In projectile motion, the motion before and after the highest point is symmetric. At the highest point, the vertical velocity becomes zero.

Given that the projectile reaches maximum height at 2 seconds, we can directly analyze the motion after the peak instead of finding initial velocity.

---

Step 2: Motion after reaching maximum height

The projectile reaches its highest point at time equal to 2 seconds.

We are asked to find the height at time equal to 3 seconds.

So, the projectile moves downward for 1 second after reaching the top.

---

Step 3: Distance fallen after the peak

At the highest point, vertical velocity is zero.

Distance fallen in downward motion depends only on gravity and time.

Distance fallen in 1 second = one half multiplied by g multiplied by time squared

Distance fallen = 0.5 multiplied by 10 multiplied by 1 multiplied by 1

Distance fallen = 5 meter

---

Step 4: Maximum height of the projectile

Time to reach maximum height is 2 seconds.

Initial upward velocity can be found using gravity symmetry:

Initial vertical velocity = g multiplied by time to reach top

Initial vertical velocity = 10 multiplied by 2

Initial vertical velocity = 20 meter per second

Maximum height reached = average velocity multiplied by time

Average velocity = (20 plus 0) divided by 2 = 10

Maximum height = 10 multiplied by 2

Maximum height = 20 meter

---

Step 5: Height at 3 seconds

Height at 3 seconds = maximum height minus distance fallen

Height at 3 seconds = 20 minus 5

Height at 3 seconds = 15 meter

---

Final Answer:

The height of the projectile at time equal to 3 seconds is
15 meter