19,170 views
29 votes
29 votes
An object is launched from the base of an incline, which is at an angle of 30° If the launch angle is 60° from the horizontal and launch speed is 10m/s. What is the total flight time ?​

User BritishSam
by
2.6k points

1 Answer

22 votes
22 votes

Answer:

4.6 s, because that is the time for the object to reach the top of the incline.

Step-by-step explanation:
that at the top of the incline, the horizontal velocity will be equal to the launch velocity, because there is no longer any force acting on the object in the horizontal direction. The only force acting on the object is gravity, which is acting in the vertical direction. So the only velocity that will be changing is the vertical velocity. The vertical velocity will be changing because of the force of gravity, which is 9.8 m/s2. The equation for the vertical velocity is: vf = vi + a*t, where vf is the final velocity, vi is the initial velocity, a is the acceleration and t is the time. The final velocity in this case is 0 m/s, because the object has reached the top of the incline and is no longer moving in the vertical direction. The initial velocity is 10 m/s, because that is the launch velocity. The acceleration is 9.8 m/s2, because that is the force of gravity. So when you plug all of those values into the equation, you get: 0 = 10 + 9.8*t and you solve for t and you get t = 4.6 s.

User Manasvi Sareen
by
3.0k points