A projectile is fired from the ground, reaches a maximum height of 26.8 m and lands...

A projectile is fired from the ground, reaches a maximum height of 26.8 m and lands a distance of 76.7 m away from the launch point. What was the projectile s launch velocity?

A) 28.2 m/s, 54.4 degrees above horizontal

B) 22.9 m/s, 27.2 degrees above horizontal

C) 16.4 m/s, 16.3 degrees above horizontal

D) 42.3 m/s, 8.2 degrees above horizontal

Expert Solution

Maximum height, h = v²sin² / 2g ...(1)
Where v is the initial velocity, is the launch angle
Horizontal range, R = v² sin2 / g ...(2)

(1) / (2) gives
h/R = (sin² / sin2) * 1/2
= [(sin*sin) / 2sin*cos)] * 1/2
= [tan]*1/2
= tan/4

tan = 4h/R
= tan^-1(4h/R)
= tan^-1(4 * 26.8 / 76.7)
= 54.4 degrees
-----------------------------------
From (1),
v = SQRT[2hg/sin²]
Substituting values,
v = SQRT[(2*26.8*9.8) / sin²(54.4)]
= 28.2 m/s

So, the correct answer is A) 28.2 m/s, 54.4 degrees above horizontal

genius_generous answered 8 months ago

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