Sol: Minimum Speed = u_{y} = Vertical Component of 15 m/s

Minimum Speed = u sin37°

$\displaystyle = 15 \times \frac{3}{5} = 9 m/s$

Time of flight $\displaystyle T = \frac{2 u sin\theta}{g} $

$\displaystyle T = \frac{2 \times 15 \times sin37^o}{10} $

$\displaystyle T = \frac{2 \times 15 \times \frac{3}{5}}{10} $

$\displaystyle T = \frac{9}{5} sec$

Horizontal Range , $\displaystyle R = \frac{u^2 sin2\theta}{g} $

$\displaystyle R = \frac{15^2 (2 sin37^o cos37^o)}{10} $

$\displaystyle R = \frac{225 (2 \times \frac{3}{5}\times \frac{4}{5})}{10} $

R = 108/5 m

Average velocity in this duration $\displaystyle v_{avg} = \frac{108/5}{9/5} = 12 m/s$

After Half time of flight projectile is at highest point , where vertical component is zero .

Initial velocity $\displaystyle \vec{u} = u cos37^o \hat{i} + u sin37^o \hat{j}$

$\displaystyle \vec{u} = 15 (4/5) \hat{i} + 15 (3/5) \hat{j}$

$\displaystyle \vec{u} = 12 \hat{i} + 9 \hat{j}$

velocity after half time of flight , $\displaystyle \vec{v} = u cos37^o \hat{i} $ ; (vertical component is zero)

$\displaystyle \vec{v} = 15 (4/5) \hat{i} $

$\displaystyle \vec{v} = 12 \hat{i} $

Change in velocity $\displaystyle = 12 \hat{i} – (12 \hat{i} + 9 \hat{j})= -9 \hat{j} $

Maximum height attained , $\displaystyle H = \frac{u^2 sin^2 \theta}{2 g} $

$\displaystyle H = \frac{15^2 sin^2 37^o}{2 \times 10} $

$\displaystyle H = \frac{225 (3/5)^2}{20} $

$\displaystyle H = \frac{225 \times 9}{20 \times 25} = \frac{81}{10}$

H ≈ 4 m