An object is launched off a 30 meter tall cliff at an angle 30 degrees below the

Question

An object is launched off a 30 meter tall cliff at an angle 30 degrees below the horizontal with an initial speed of 10.0 m/s. Determine first the x- and y- components of its initial velocity.

a. x-component:  

Answer:________________ m/s.

b. y-component:

Answer:_________________ m/s

c. Now determine the time it would take for the projectile to land.

time to land: Answer:______________ s

d. and determine the distance along the ground from the cliff edge to the point that the projectile actually hits.

distance along ground Answer:_______________ m

e. Determine the time and position at which the projectile is at a height of 15 meters.

time at which projectile is at a height of 15 m Answer:____________ s

f. Position, as measured along the ground from the base of the cliff, at which the projectile is at a height of 15m Answer:___________ m

g. Determine the y-component of the velocity of the projectile at the moment of impact.

y-component of velocity at moment of impact:  Answer:_____________m/s

h. Now find the speed of the projectile at the moment of impact using both the x-and y-components of its final velocity.

Speed of projectile at moment of impact: Answer:____________ m/s

And, finally, please determine the angle of the projectile's final trajectory as it hits the ground. Please find the angle as measured from the vertical direction. As in the previous problem, please use the x- and y-components of its final velocity to do this.

i. Angle of the projectile's path, as measured from the vertical direction, at the moment of impact:

Answer:__________degrees.

Explanation / Answer

a, v0x = 10 cos30 = 8.66 m/s

(b) v0y = 10 sin30 = 5 m/s

(c) y = v0y t + ay t^2 / 2

0 - 30 = 5 t - 9.8 t^2 / 2

- 4.9 t^2 + 5 t + 30 = 0

t = 3.04 sec  

(D) x = v0x t = 8.66 x 3.04 = 26.3 m

(E) at highets vfy = 0

vfy^2 - v0y^2 = 2 ay h

h = 1.28 m

from ground, h_max = 31.28 m

f. t = v0y / g = 0.51 sec  

g. 15 - 30 = 5 t - 4.9 t^2

t = 2.33 sec  

h. x = 8.66 x 2.33 = 20.2 m

i. vy = 5 - (9.8 x 3.04) = - 24.8 m/s

j. vy = - 24.8 m/s  

vx = 8.66 m/s

v = sqrt( vy^2 + vx^2) = 26.3 m/s

k theta = tan^-1(vx / vy) = 19.2 deg

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