A projectile is launched from the top of a cliff that is 65 m above ground level

Question

A projectile is launched from the top of a cliff that is 65 m above ground level. The projectile is launched with an initial speed of 27 m/s at an angle of 36 degree below the horizontal. When the horizontal distance of the projectile from the cliff is 42 m. answer the following questions. How high above the ground is the projectile? What are the horizontal and vertical components of the velocity of the projectile? What are the magnitude and direction of the velocity of the projectile? What are the magnitude and direction of the acceleration of the projectile?

Explanation / Answer

here,

12)

height , h = 65 m

initial speed , u = 27 m/s

theta = 36 degree

when x = 42 m

let the time taken be t

x = u * cos(theta) * t

42 = 27 * cos(36) * t

t = 1.92 s

a)

the height of projectile above ground , h' = h- ( u * sin(theta) * t + 0.5 * g * t^2 )

h' = 65 - ( 27 * sin(36) * 1.92 + 0.5 * 9.81 * 1.92^2) m

h' = 16.45 m

b)

the horizontal component of velocity , vx = u * cos(theta) = 21.8 m/s

the vertical component of velocity , vy = u * sin(theta) + g * t

vy = 27 * sin(36) + 9.81 * 1.92 m/s

vy = 34.7 m/s

c)

the magnitude , |v| = sqrt(vx^2 + vy^2)

|v| = sqrt(21.8^2 + 34.7^2) m/s

|v| = 40.98 m/s

direction , theta = arctan(34.7/21.8)

theta = 57.9 degree below the horizontal

d)

the accelration is 9.81 m/s^2 acting downwards

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