physics question

Question

so... the answer is t=2.04 and y=20.40 but idk how to do this:

An object is projected upward with a 30 degree launch angle and an initial speed of 40m/s. How long will it take for the object to reach to top of its trajectory? How high is this?

Answer 1

Well we know that v₀ = 40 m/s and if the launch angle is 30 degrees, we can determine what the velocity in the x and y direction are:

v_0x = 40cos30 = 34.64 m/s

v_0y = 40sin30 = 20 m/s

The object will reach the top of its trajectory when v_y = 0 m/s because it will no longer be traveling upwards. Therefore we can use the following equation to solve for how high up it goes:

v_y² = v_0y² + 2gΔy

0² = 20² + 2(-9.81)Δy

0 = 400 - 19.62Δy

19.62Δy = 400

Δy = 400/19.62 = 20.38 m

We can use this information to solve for t now with the following equation:

Δy = 1/2(v_y + v_0y)t

20.38 = 1/2(0 + 20)t

20.38 = 10t

t = 20.38/10 = 2.04 s