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junior college 1 | H1 Maths
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Hello! can someone kindly help me with this question? I keep getting values that are higher than the answer provided and idk what went wrong. Thank u for ur effort!
Date Posted: 4 years ago
Views: 325
What value are you using for your gravitational acceleration?
(a)
u = 3.25 m/s (upwards)
v₁ = 0 m/s
a = -9.81 m/s²
t = 3.25/9.81 = 0.3313 s
time taken upwards = 0.3313 s
v₁ = 0 m/s
v₂ = -3.25 m/s
a = -9.81 m/s²
t = 3.25/9.81 = 0.3313 s
time taken downwards = 0.3313 s back to original "drop-off" pt
time taken to reach the ground
= 8.75 - 0.3313 - 0.3313
= 8.0874 s
x= 3.25 x 8.0874 + 0.5 . 9.81 . (8.8074)²
x = 347.1 m
u = 3.25 m/s (upwards)
v₁ = 0 m/s
a = -9.81 m/s²
t = 3.25/9.81 = 0.3313 s
time taken upwards = 0.3313 s
v₁ = 0 m/s
v₂ = -3.25 m/s
a = -9.81 m/s²
t = 3.25/9.81 = 0.3313 s
time taken downwards = 0.3313 s back to original "drop-off" pt
time taken to reach the ground
= 8.75 - 0.3313 - 0.3313
= 8.0874 s
x= 3.25 x 8.0874 + 0.5 . 9.81 . (8.8074)²
x = 347.1 m
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Hope this helps! Remember to take note of the signs for velocity as it is a vector quantity.
Date Posted:
4 years ago
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