Question

The maximum height reached by a projectile is 125m .Its time of flight is

Answer 1

Answer:

• Time of Flight (T) = 10 Seconds.

Given:

1. Maximum height (H) = 125 Meters.
2. Acceleration due to gravity (g) = 10 m/s².

Explanation:

$\rule{300}{1.5}$

From Maximum height Formula,

We Know,

$\large{\boxed{\bold{H = \dfrac{u^2 Sin^2 \theta}{2g}}}}$

$\bold{Here}\begin{cases}\text{u denotes Initial velocity} \\ \theta \: \text{Denotes Angle} \\ \text{g denotes Acceleration due to gravity}\end{cases}$

$\boxed{\setlength{\unitlength}{1 cm}\thicklines\begin{picture}(6.65,3.2) \put(0.3,1){\line(1,0){6}} \qbezier(0.3,1)(3,3)(6.3,1)\put(0.3,1){\vector(1,1){1}} \qbezier(0.5,1.2)(0.65,1,1)(0.6,1)\put(0.77,1.065){ \theta }\put(0.5,1.8){u}\put(3.15,1.5){\vector(0,1){0.5}}\put(3.15,1.5){\vector(0,-1){0.5}}\put(3.3,1.4){H}\put(4.3,0.7){\vector(-1,0){4}}\put(4.3,0.7){\vector(1,0){2}}\put(3.3,0.3){R}\end{picture}}$

Now, From the Formula,

$\large{\boxed{\tt H = \dfrac{u^2 Sin^2 \theta}{2g}}}$

Substituting the values,

$\large{\tt \hookrightarrow 125 = \dfrac{u^2 Sin^2 \theta}{2 \times 10}}$

$\large{\tt \hookrightarrow 125 = \dfrac{u^2 Sin^2 \theta}{20}}$

Cross - Multiplying,

$\large{\tt \hookrightarrow 125 \times 20 = u^2 Sin^2 \theta}$

$\large{\tt \hookrightarrow 2500 = u^2 Sin^2 \theta}$

$\large{\tt \hookrightarrow u^2 Sin^2 \theta = 2500}$

$\large{\tt \hookrightarrow (u Sin \theta)^2 = 2500}$

$\large{\tt \hookrightarrow u Sin \theta = \sqrt{2500}}$

$\large{\boxed{\tt u Sin \theta = 50}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

From Time of Flight Formula,

We have,

$\large{\boxed{\bold{T = \dfrac{2u Sin \theta}{g}}}}$

$\bold{Here}\begin{cases}\text{u denotes Initial velocity} \\ \theta \: \text{Denotes Angle} \\ \text{g denotes Acceleration due to gravity}\end{cases}$

Now,

$\large{\boxed{\tt T = \dfrac{2u Sin \theta}{g}}}$

Substituting the values,

$\large{\tt \hookrightarrow T = \dfrac{2 \times 50}{10}}$

• ∵u sinθ = 50 m/s.

$\large{\tt \hookrightarrow T = \dfrac{2 \times 50}{10}}$

$\large{\tt \hookrightarrow T = \dfrac{2 \times \cancel{50}}{\cancel{10}}}$

$\large{\tt \hookrightarrow T = 2 \times 5}$

$\huge{\boxed{\boxed{\tt T = 10 \: Sec}}}$

∴ Time of Flight is 10 Seconds.

$\rule{300}{1.5}$

Answer 2

◆【Solution】◆

● Max height ●

When a body projected at any angle than the horizontal component of velocity keep on its magnitude because there is no acceleration at horizontal level.

But vertical component keep on decreasing due to acceleration. due to gravity. After a certain period i.e half of time period it become zero and displacement in y direction in this time is know as MAXIMUM HEIGHT OF PROJECTILE.

♂ Mathematically ♂

Max height = ◆【u²Sin²Ѳ/2g】◆

Ѳ is the angle of projection.

u is the velocity of projection

acc to you it is 125m

u²Sin²Ѳ/2g = 125

u²Sin²Ѳ = 2500

{usinѲ = 50m/sec......vertical velocity}

Displacement in y direction = 125m

Initial velocity = 50m/sec

Time taken = ?

V = U - gT

0 = 50 - 10T

◆【TIME TAKEN = 5sec】◆

#answerwithquality

#BAL