Question

An object is launched at a velocity of 20 m/s in a direction making at an angle of 25⁰ upward with the horizontal. Find

(a) The maximum height reached by the object.

(b) The magnitude of the velocity of the object before it hits the ground.

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Answer 1

below

Step-by-step explanation:

Vertical position = vo t + 1/2 at ^2

where vo = 20 sin25 =8.45 and a = -9.81 m/s^2

Vertical position = - 4.905 t^2 + 8.45 t

max will occur at t = -b/2a = -8.45 / (2* -4.905) = .861 s

Plug in this value to find vertical height

vertical position = 8.45(.861) - 1/2 (9.81)(.861)^2 = 6.92 m

b) magnitude when it comes down = magnitude when it went up = 20 m/s