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wassup
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Ok, I thought I knew how to do this, but for some reason, according to my homework website, my answer is wrong. So I'm turning to this board before I beat my head against the wall. Please help me.
A ball is thrown with an initial speed of 31 m/s at an angle of 45�. The ball is thrown from a height of 12 m and lands on the ground.
Find the time of flight.
Initial vertical upward velocity = 31 sin 45 = 26.378m/s.
The displacement [Yf – Yo] = –12 m
Yf -Yo = Vot - 0.5*g*t^2
–12 = 26.378 t – 4.905 *t^2
4.905 *t^2 – 26.378 t – 12 = 0
--------------------------------------...
Using b^2 – 4ac = 26.378 ^2– 4*4.905*(– 12)
√ (b^2 – 4ac) = 30.516
– b ±√ (b^2 – 4ac) = 26.378 ± 30.516
= 56.894 leaving the negative time value
Dividing by 2a = 9.81
t = 5.7996 s
Is my work right?
Homework Statement
A ball is thrown with an initial speed of 31 m/s at an angle of 45�. The ball is thrown from a height of 12 m and lands on the ground.
Find the time of flight.
The Attempt at a Solution
Initial vertical upward velocity = 31 sin 45 = 26.378m/s.
The displacement [Yf – Yo] = –12 m
Yf -Yo = Vot - 0.5*g*t^2
–12 = 26.378 t – 4.905 *t^2
4.905 *t^2 – 26.378 t – 12 = 0
--------------------------------------...
Using b^2 – 4ac = 26.378 ^2– 4*4.905*(– 12)
√ (b^2 – 4ac) = 30.516
– b ±√ (b^2 – 4ac) = 26.378 ± 30.516
= 56.894 leaving the negative time value
Dividing by 2a = 9.81
t = 5.7996 s
Is my work right?