Question

A projectile is fired with an initial speed of 48.6 m/sat an angle of 44.2 ∘ above the horizontal on a long flat firing range.

Part A

Determine the maximum height reached by the projectile.

ymax =

nothing

  m  

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Part B

Determine the total time in the air.

t =

nothing

  s  

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Part C

Determine the total horizontal distance covered (that is, the range).

Δx =

nothing

  m  

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Part D

Determine the speed of the projectile 1.30 s after firing.

v =

nothing

  m/s  

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Part E

Determine the direction of the motion of the projectile 1.30 s after firing.

θ =

nothing

  ∘ above the horizontal

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Answer 1

here,

the initial speed , u= 48.6 m/s

theta = 44.2 degree

a)

the maximum height reached by the projectile , hm = (u * sin(theta))^2 /( 2g)

hm = ( 48.6 * sin(44.2))^2 /(2*9.81)

hm = 58.5 m

b)

the total time in air , t = (2 * u * sin(theta))/g

t = 2 * 48.6 * sin(44.2)/9.81

t = 6.91 s

c)

the total horizontal distance covered , R = u * cos(theta) * t

R = 48.6 * cos(44) * 6.91 m = 241.6 m

d)

t1 = 1.3 s

the vertical speed , vy = u * sin(theta) - g * t1

vy = 48.6 * sin(44.2) - 9.81 * 1.3 m/s = 21.13 m/s

the horizontal speed , vx = u * cos(theta) = 34.96 m/s

the speed of projectile , v = sqrt(vx^2 + vy^2)

v = sqrt(34.96^2 + 21.13^2) = 40.85 m/s

e)

the direction of motion , theta = arctan(21.13 /34.96) = 31.1 degree above the horizontal