A projectile is fired with an initial speed of 28 m/s at an angle of 62° above t

Question

A projectile is fired with an initial speed of 28 m/s at an angle of 62° above the horizontal. The object hits the ground 7.1 s later. (Neglect air (a) How much higher or lower is the launch point relative to the point where the projectile hits the ground? (If the point where the projectile hits the ground is lower, enter a negative number.) (b) To what maximum height above the launch point does the projectile rise? (c) What are the magnitude and direction of the projectile's velocity at the instant it hits the ground? magnitude direction m/s (counterclockwise from the +x-axis)

Explanation / Answer

t = 7.1 s

y = usin62*t-1/2gt^2

y = 28sin62*7.1-1/2*9.8*7.1^2 = -71.47 m

b) h max = u^2sin^262/2g = 28^2*sin^62/2*9.8 = 31.18 m

c) vx = 28cos62 = 13.14 m/s

vy = 28sin62-9.8*7.1 = -44.85 m/s

magnitude v = sqrt(vx^2+vy^2) = 46.7 m/s

theta = tan^-2(-44.85/13.14) = 286.3 from positive x axis

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