A projectile is launched from a platform 10 feet above the ground at an angle of $50^{\circ}$ from the ground. If the initial velocity is $130 \mathrm{ft} / \mathrm{s}$, for what time interval is the projectile at least 60 feet above the ground? The height, in feet, of a projectile launched at an angle of $\theta$ from the ground with an initial velocity of $\mathrm{v}_{0}(\mathrm{ft} / \mathrm{s}$ ) and a starting height of $\mathrm{s}$ (feet) is given by the formula $y=-16 t^{2}+v_{0} \cdot \sin (\theta) t+s$. Use the formula, along with $\sin \left(50^{\circ}\right)=0.766$ to solve the inequality. Round to the nearest tenth of a second if necessary.
A) 0.5 to 5.6 seconds
B) 0.8 to 6.3 seconds
C) 0.9 to 7.1 seconds
D) 0.6 to 5.7 seconds

Question

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The Answer of A projectile is launched from a platform 10 feet above the ground at an angle of $50^{\circ}$ from the ground. If the initial velocity is $130 \mathrm{ft} / \mathrm{s}$, for what time interval is the projectile at least 60 feet above the ground? The height, in feet, of a projectile launched at an angle of $\theta$ from the ground with an initial velocity of $\mathrm{v}_{0}(\mathrm{ft} / \mathrm{s}$ ) and a starting height of $\mathrm{s}$ (feet) is given by the formula $y=-16 t^{2}+v_{0} \cdot \sin (\theta) t+s$. Use the formula, along with $\sin \left(50^{\circ}\right)=0.766$ to solve the inequality. Round to the nearest tenth of a second if necessary.
A) 0.5 to 5.6 seconds
B) 0.8 to 6.3 seconds
C) 0.9 to 7.1 seconds
D) 0.6 to 5.7 seconds

A Projectile Is Launched from a Platform 10 Feet Above 50∘50^{\circ}50∘

A Projectile Is Launched from a Platform 10 Feet Above $50^{\circ}$