An arrow is shot from a height of 1.95 m toward a cliff of height h. it is shot with a velocity of 29 m/s at an angle of 60â° above the horizontal. it lands on the top edge of the cliff 3.8 s later.

a) what is the height of the cliff (in m)?

(b) what is the maximum height (in m) reached by the arrow along its trajectory?

(c) what is the arrow's impact speed (in m/s) just before hitting the cliff?

Answer: it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.explanation:   we have equation of motion ,where s is the displacement, a is the acceleration, u is the initial velocity and t is the time taken.average velocity = displacement/ time taken.average velocity = 10.69 miles/hr = 10.69*1.6*1000/(60*60) = 4.75 m/s4.75 = s/ts = 4. 1we also have acceleration = 2.18 m/s², initial velocity = 0 m/sso it will take   4.36 seconds to reach her average velocity 10.69 miles/hour.