physics question?

Question

An object is projected upward with a 30° launch angle from the ground and an initial speed of 60 m/s. For how many seconds will it be in the air? How far will it travel horizontally? Assume it returns to its original height.

answer's t=6.12 and x=317.18; i don't know how to do this problem

Answer 1

Let

α = 30° be the launch angle,

v = 60 m/s be the initial speed,

g = 9.8 m/s² be gravitational acceleration,

t be time variable, starting with t = 0 at the moment of launch.

The object follows a parabola resulting from a combination of two movements:

- A straight line at angle α with constant speed v due to inertia, which is what the object would

follow in the absence of gravity.

- A vertical fall due to gravity, which is what the object would follow

in the absence of any initial velocity.

You can view the trajectory in an X-Y coordinate space, where

the x coordinate is along the ground, and

the y coordinate is vertical, with the origin at the point of launch.

Then [x(t), y(t)] is the position of the object after time t.

The horizontal distance x(t) is unaffected by the vertical fall, therefore

x(t) = v t cos(α)

In contrast, the vertical distance y(t) is a combination of the two movements:

y(t) = v t sin(α) - g t²/2.

The portion v t sin(α) is the height the object would reach after time t in the absence of gravity, and

g t²/2 is the amount of fall the object would suffer without any initial velocity.

The flight continues until the object hits the ground at some time t₁,

which we can calculate because at time t₁, y(t₁) = 0. That is,

v t₁ sin(α) - g t₁²/2 = 0

Therefore

t₁ = 2 v sin(α) / g

Having obtained t₁, we can calculate how far the object got horizontally by:

x(t₁) = v t₁ cos(α)

Substituting actual values:

t₁ = 2 . 60 m/s . sin(30°) / (9.8 m/s²) = 6.12 s

x(t₁) = 60 m/s . 6.12 s . cos(30°) = 318 m

I assume that the difference between your 317.18 m and my 318 m is due to different rounding.