A projectile is fired with an initial speed of 29 m/s at an angle of 66° above t

Question

A projectile is fired with an initial speed of 29 m/s at an angle of 66° above the horizontal. The object hits the ground 7.7 s later. (Neglect air resistance.) (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

here,

initial speed , u = 29 m/s

theta = 66 degree

time of flight , t = 7.7 s

a)

let the height relative to the ground be h

h = u * sin(theta) * t - 0.5 * g * t^2

h = 29 * sin(66) * 7.7 - 0.5 * 9.8 * 7.7^2 m

h = - 86.5 m

so the launch point is 86.5 m above the ground

b)

the maximum height above the launch point , Hmax = (u * sin(theta))^2 /2g

Hmax = ( 29 * sin(66))^2 /(2*9.81) m

Hmax = 35.8 m

c)

the final vertical velocity , vy = u * sin(theta) - g * t

vy = 29 * sin(66) - 9.81 * 7.7 m/s

vy = - 49 m/s

final horizontal velocity ,vx = u * cos(theta) = 29 * cos(66) = 11.8 m/s

the magnitude of velocity , |v| = sqrt(vx^2 + vy^2) = 50.4 m/s

theta = arctan(vy/vx) = 76.5 degree below the horizontal

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