Question #12166

A projectile is thrown with the speed of 25 meter per seconds in a direction 30° above the horizontal.
-how long does it take to reach the maximum height.
-how long does it reach the Projectile along in the air.
-how far does it go.

Expert's answer

Here is the formula for the height of a projectile:

H(t) = Vt·sinα - gt²/2

Let's derivate it by t:

H'(t) = V·sinα - gt

V·sinα - gt = 0 ==> t = V·sinα/g = 25·sin30°/9.8 ≈ 1.2755 s.

-how long does it reach the Projectile along in the air.

The time of a flight is the time needed to reach the maximum height doubled:

T = 2t ≈ 2·1.2755 = 2.5510.

-how far does it go.

Using the obtained flight time:

L = VT·cosα ≈ 25*2.5510*cos30 = 55.2308 m.

H(t) = Vt·sinα - gt²/2

Let's derivate it by t:

H'(t) = V·sinα - gt

V·sinα - gt = 0 ==> t = V·sinα/g = 25·sin30°/9.8 ≈ 1.2755 s.

-how long does it reach the Projectile along in the air.

The time of a flight is the time needed to reach the maximum height doubled:

T = 2t ≈ 2·1.2755 = 2.5510.

-how far does it go.

Using the obtained flight time:

L = VT·cosα ≈ 25*2.5510*cos30 = 55.2308 m.

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