A bullet is fired with an initial velocity of 100 m/s at an angle of 53° from the level ground. Determine the amount of time the bullet spends in the air.
(Use 10 m/s for gravity)
2 answers
*10 m/s ^2
Vertical velocity:
Vy = V₀ sin θ = 100 ∙ sin 53° = 79.8636
Height:
h = Vy ∙ t - g t² / 2
When the bullet touches the ground the height will be zero.
Vy ∙ t - g t² / 2 = 0
Add g t² / 2 to both sides
Vy ∙ t = g t² / 2
Divide both sides by t
Vy = g t / 2
Multiply both sides by 2
2 Vy = g t
2 Vy / g = t
t = 2 Vy / g
t = 2 ∙ 79.8636 / 10 = 15.97272
t ≈ 16 sec
Vy = V₀ sin θ = 100 ∙ sin 53° = 79.8636
Height:
h = Vy ∙ t - g t² / 2
When the bullet touches the ground the height will be zero.
Vy ∙ t - g t² / 2 = 0
Add g t² / 2 to both sides
Vy ∙ t = g t² / 2
Divide both sides by t
Vy = g t / 2
Multiply both sides by 2
2 Vy = g t
2 Vy / g = t
t = 2 Vy / g
t = 2 ∙ 79.8636 / 10 = 15.97272
t ≈ 16 sec
1 answer