Asked by Tshepo success
A ball is thrown at 21m/s at 30degree above the horizontal from the top of a roof 16m high.calculate the time of flight
Answers
Answered by
drwls
The initial vertical velocity component is
Vyo = 21 sin 30 = 10.5 m/s.
Height above ground (y) is given by the equation:
y = y(0) + Vyo*t - (g/2)t^2
= 15 + 10.5 t - 4.9 t^2
Set y = 0 and solve for t. Take the positive one of the two roots of the quadratic equation.
4.9 t^2 -10.5 t -10 = 0
t = (1/9.8)*[10.5 + sqrt(110.25 +196)]
= 2.86 seconds
Vyo = 21 sin 30 = 10.5 m/s.
Height above ground (y) is given by the equation:
y = y(0) + Vyo*t - (g/2)t^2
= 15 + 10.5 t - 4.9 t^2
Set y = 0 and solve for t. Take the positive one of the two roots of the quadratic equation.
4.9 t^2 -10.5 t -10 = 0
t = (1/9.8)*[10.5 + sqrt(110.25 +196)]
= 2.86 seconds
Answered by
Anonymous
find range
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