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.
1
Expert's answer
2012-08-09T10:49:49-0400
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.
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