Question

A body projected from base of an inclined plane with the velocity 45 m/s. where the angle of inclination is 37⁰. If the angle formed by projectile is 45°, find the maximum height, Range, Time of flight.

⚠️⚠️⚠️irrelevant answers are reported⚠️⚠️⚠️​

Answer 1

Solution:

Given:

• u = 45 m/s
• θ = 37°
• α = 45°

To find:

• Maximum Height
• Range
• Time of Flight

Solution:

For finding Maximum Height:

$\tt{H = \dfrac{u^2sin^2\alpha}{2g cos \theta}}$

$\Rightarrow \: \sf{H = \dfrac{(45)^2sin^{2}45}{2 \times 10 \times cos37^{\circ}}}$

$\Rightarrow \: \sf{H=\dfrac{45 \times 45 \times \dfrac{1}{2}}{2 \times 10 \times \dfrac{4}{5}}}$

$\Rightarrow \: \sf{H = \dfrac{1012.5}{16}}$

$\therefore \: \boxed{\bf{H = 63.28 \: m}}$

For finding Range:

$\tt{R = \dfrac{u^{2}sin2\alpha}{g\:cos\theta}-\bigg(\dfrac{2u^{2}sin^{2}\alpha}{g \: cos \theta}\times tan\theta\bigg)}$

$\dashrightarrow \: \sf{R = \dfrac{45^{2}sin[2\times45^{\circ}]}{10 \times \dfrac{4}{5}}-\bigg(\dfrac{2(45)^2\times\dfrac{1}{2}}{10 \times \dfrac{4}{5}}\times tan 37^{\circ}\bigg)}$

$\dashrightarrow \: \sf{R = \dfrac{45 \times 45}{8}-\bigg(\dfrac{45\times45\times3}{8\times4}\bigg)}$

$\dashrightarrow \: \sf{R=253-189}$

$\therefore \: \boxed{\bf{R = 64 \: m}}$

For finding Time of Flight:

$\tt{T=\dfrac{2u\:sin\alpha}{g \: cos \theta}}$

$\implies \sf{T = \dfrac{2(45)sin45^{\circ}}{10 \: cos 37^{\circ}}}$

$\implies \sf{T=\dfrac{2(45)\times \dfrac{1}{\sqrt{2}}}{10\times\dfrac{4}{5}}}$

$\implies \sf{T=\dfrac{45 \sqrt{2}}{8}}$

$\therefore \boxed{\bf{T \approx 8\:sec}}$