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A Projectile Is Launched from a Platform 10 Feet Above 5050^{\circ}

Question 124

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A projectile is launched from a platform 10 feet above the ground at an angle of 5050^{\circ} from the ground. If the initial velocity is 130ft/s130 \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 θ\theta from the ground with an initial velocity of v0(ft/s\mathrm{v}_{0}(\mathrm{ft} / \mathrm{s} ) and a starting height of s\mathrm{s} (feet) is given by the formula y=16t2+v0sin(θ) t+sy=-16 t^{2}+v_{0} \cdot \sin (\theta) t+s . Use the formula, along with sin(50) =0.766\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

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