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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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`
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:
By the condition of the problem, `(2usinalpha)/(g)=4.08` --- (i)
And, `(ucosalpha*2usinalpha)/(g) =76.2 ` --- (ii)
`ucosalpha=76.2/4.08=18.67647 ` --- (iii)
`rArr usinalpha=(76.2*9.81)/(2*18.67647)=20.01` --- (iv)
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