Question

a projectile is fired from ground with a velocity of 20m/s with an angle of 60° with the horizontal. Find time of flight, h.max,range

Answer 1

time of flight =2usintheeta\g
2*20 sin60°/10
2√3is the time

hmax=u²/2g
=400/20
=20
range= u²sin2theeta/g
=400sin120°/10
=20√3
hope this helps you!!

Answer 2

Answer:

The time of flight, maximum height, and range of the projectile is $2\sqrt{3} \ seconds$, 15m, $20\sqrt{3} m$ respectively.

Explanation:

From the question we have,

The velocity of projection(u)=20m/s

The angle of projection(θ)=60°

Time of flight

$T=\frac{2usin\theta}{g}$        (1)

Where,

T=time of flight

u=velocity of projection

θ=angle of projection

g=acceleration due to gravity=10m/s²

By placing the values in equation (1) we get;

$T=\frac{2\times 20\times sin60\textdegree}{10}$

$T=4\times \frac{\sqrt{3} }{2}$

$T=2\sqrt{3} \ seconds$      (2)

Maximum height

$H_m_a_x=\frac{u^2sin^2\theta}{2g}$      (3)

$H_m_a_x$=maximum height attained by the projectile

By placing all the required values in equation (3) we get;

$H_m_a_x=\frac{(20)^2\times sin^260\textdegree}{2\times 10}$

$H_m_a_x=20\times( \frac{\sqrt{3} }{2} )^2$

$H_m_a_x=20\times \frac{3}{4}$

$H_m_a_x=5\times 3$

$H_m_a_x=15m$     (4)

Range of projectile

$R=\frac{u^2sin2\theta}{g}$      (5)

R=range of the projectile

By placing all the required values in equation (5) we get;

$R=\frac{(20)^2\times sin2(60\textdegree)}{10}$

$R=\frac{(20)^2\times sin120\textdegree}{10}$

$R=20\times 2\times \frac{\sqrt{3} }{2}$

$R=20\sqrt{3} m$   (6)

Hence, the time of flight, maximum height, and range of the projectile is $2\sqrt{3} \ seconds$, 15m, $20\sqrt{3} m$ respectively.

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