Question

A projectile returns to its original height 4.08 s after being launched, during which time it travels 76.2 m horizontally. If air resistance can be neglected, what was the projectile's initial speed?

Answer 1

Answer:27.35 m/s

Explanation:

Given

Time of Flight of Projectile T=4.08 s

Range of Projectile =76.2 m

Time Of Flight of Projectile is given by

$T=\frac{2u\sin \theta }{g}----------1$

where u=initial Velocity

$\theta =$Launch angle

g=acceleration due to gravity

Range is given by $R=\frac{u^2\sin 2\theta }{g}------2$

divide 1 and 2

$\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }$

$\frac{R}{T}=u\cos \theta ------3$

$u\sin \theta =\frac{Tg}{2}------4$

squaring and adding 3 & 4 we get

$u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2$

$u^2=(19.992)^2+(18.676)^2$

$u=\sqrt{748.49}$

$u=27.35 m/s$                              

Answer 2

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

The initial speed of the projectile is 27.35 m/s.

What is the initial speed?

The initial speed is the speed of the object at the beginning of the measurement or the starting speed.

Given information-

The time taken by projectile to returns its original height is 4.08 s.

The distance traveled by it is 76.2 m.

Air resistance can be neglected.

The time of flight of a projectile motion can be given as,

$T=\dfrac{2u\sin \theta}{g}$

Let the above equation is equation 1.

Here, $u$ is the initial velocity, and $\theta$ is the angle of launch.

Rewrite the above equation as,

$u\sin \theta =\dfrac{Tg}{2}$  

Let the above equation is equation 2.

Now the range of the projectile motion can be given as,

$R=\dfrac{u^2\sin (2\theta) }{g}\\R=\dfrac{u^22\cos \theta\sin \theta }{g}$

Divide this equation by the equation 1 as,

$\dfrac{R}{T}=u\cos \theta$

Square and add the the above equation and equation 2 as,

$u^2(\cos^2 \theta+\sin^2 \theta)=\dfrac{R}{T}+\dfrac{Tg}{2}\\u^2(1)=\dfrac{76.2}{4.08}+\dfrac{4.08\times9.81}{2}\\u=27.35\rm m/s$

Hence, the initial speed of the projectile is 27.35 m/s.

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