1) A ball is thrown at an angle of 56.4° above thehorizontal with an initial vel

Question

1) A ball is thrown at an angle of 56.4° above thehorizontal with an initial velocity of 31.8 m/s from the top of acliff 61.2 m high.
What is the time of flight? :
What horizontal distance did the ball travel from the base of thecliff? 2)A stone is thrown horizontally with an initial speedof 8.0 m/s from the edge of a cliff. A stop watch measures thestone's time of flight from the top of the cliff to the bottom tobe 3.8 s. What is the height of the cliff?
1) A ball is thrown at an angle of 56.4° above thehorizontal with an initial velocity of 31.8 m/s from the top of acliff 61.2 m high.
What is the time of flight? :
What horizontal distance did the ball travel from the base of thecliff? 2)A stone is thrown horizontally with an initial speedof 8.0 m/s from the edge of a cliff. A stop watch measures thestone's time of flight from the top of the cliff to the bottom tobe 3.8 s. What is the height of the cliff?

Explanation / Answer

1) the horizontal and vertical components of initial velocityare u = 31.8 cos(56.4) = 17.6 m/s, v = 31.8 sin(56.4) = 26.5 m/s time of flight can be computed from the vertical equation of motionas h = - 61.2 = vt - (1/2)gt^2 = 26.5t - 4.9t^2 t = 7.15 s (the other solution is negative). horizontal distance = ut = 17.6*7.15 = 125.8 m 2) in this case u=8 and v=0 h = (1/2)gt^2 = .5*9.8*3.8^2 = 70.8 m

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.