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

Question

An object is launched off a 30 meter tall cliff at an angle 30 degrees above 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 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. Also, determine the highest point that the projectile reaches (as measured from the ground).

Projectile's highest point:

Answer:___________ m

f. and determine the time at which the projectile reaches this highest point.

time of highest point:  Answer:________ s

g. 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

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

i. 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

j. 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

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

Angle of the projectile's path, as measured from vertical, at 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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