Question

For the following question, enter the correct numerical value up to TWO decimal places. If the numerical value has more than two decimal places, round-off the value to TWO decimal places. (For example: Numerical value 5 will be written as 5.00 and 2.346 will be written as 2.35) A particle is thrown with velocity \( u \) making angle \( \theta \) with vertical. It just crosses the top of two poles each of height \( h \) after 1 s and 3 s respectively. The maximum height of the projectile is _____

Hint:

If a projectile crosses the same height at times \( t_1 \) and \( t_2 \), then \( u_y = \dfrac{g(t_1+t_2)}{2} \).

Answer 1

Correct Answer: 19.6

Step 1: Vertical motion of the projectile is given by \[ y = u_y t - \frac{1}{2}gt^2, \] where \( u_y \) is the vertical component of velocity.
Step 2: Since the projectile crosses the same height \( h \) at times \[ t_1 = 1\,\text{s}, \quad t_2 = 3\,\text{s}, \] the vertical component of velocity is \[ u_y = \frac{g(t_1+t_2)}{2}. \]
Step 3: Substituting values, \[ u_y = \frac{9.8(1+3)}{2} = 19.6\,\text{m s}^{-1}. \]
Step 4: Maximum height of a projectile is \[ H = \frac{u_y^2}{2g}. \] \[ H = \frac{(19.6)^2}{2 \times 9.8} = 19.6\,\text{m}. \] Hence, the maximum height of the projectile is \[ \boxed{19.60} \]