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

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

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Asked on by Kelsia2

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llltkl | College Teacher | (Level 3) Valedictorian

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Let the angle of launch of the projectile, with respect to the horizontal be `alpha` and the initial firing velocity, u

Constant horizontal velocity: `ucosalpha`

Initial vertical velocity:` usinalpha`

Time to reach the peak: t

From the laws of motion: `0=usinalpha-g*t`

`rArr t=usinalpha/g`

By symmetry of the parabolic motion, time of ascent = time of descent.

So, total time of its flight=`(2usinalpha)/g`

Horizontal distance covered in this time:

`(ucosalpha*2usinalpha)/(g) =(u^2sin2alpha)/g`

By the condition of the problem, `(2usinalpha)/(g)=4.08` --- (i)

And, `(ucosalpha*2usinalpha)/(g) =76.2 ` --- (ii)

(ii)/(i)

`ucosalpha=76.2/4.08=18.67647 ` --- (iii)

Again, (ii)/(iii)

`2usinalpha/g=76.2/18.67647`

`rArr usinalpha=(76.2*9.81)/(2*18.67647)=20.01` --- (iv)

(iv)/(iii)

`tanalpha=20.01/18.67647=1.07`

`alpha=arctan(1.07)=47` degrees

From (iv),

`u=20.01/sinalpha=20.01/0.73=27.37` m/s

Sources:

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