Level:
Project ID:
9000022901
Accepted:
1
Clonable:
0
Easy:
0
An arrow has been shot at the angle
\(60^{\circ }\) at the
speed \(10\, \mathrm{m}\, \mathrm{s}^{-1}\).
Find the time when the height equals to the horizontal distance from the take-off
point. Hint: The position is given by the equations
\(x = v_{0}t\cdot \cos \alpha \),
\(y = v_{0}t\cdot \sin \alpha -\frac{1}
{2}gt^{2}\). Use
\(g = 10\, \mathrm{m}\, \mathrm{s}^{-2}\) as an
acceleration of gravity.
\(\left (\sqrt{3} - 1\right )\, \mathrm{s}\)
\(\left (\sqrt{3} + 1\right )\, \mathrm{s}\)
\(\sqrt{3}\, \mathrm{s}\)
\(\left (\sqrt{2} - 1\right )\, \mathrm{s}\)
\(\left (\sqrt{2} + 1\right )\, \mathrm{s}\)