Question #145252
An Object is thrown into space with a velocity of 15 m/s at an angle of 60o with
respect to the horizontal, Calculate (i) maximum height
(ii) time of flight

(iii) horizontal range. (Given g = 9.8 m/s-2
)
1
Expert's answer
2020-11-20T09:30:32-0500

See https://courses.lumenlearning.com/boundless-physics/chapter/projectile-motion/ for more details.


1. The maximum height for a body launched at an angle to the horizontal, is:


hmax=v2sin2θ2gh_{max} = \dfrac{v^2\sin^2\theta}{2g}

where v=15m/sv=15m/s is the initial speed of the body, θ=60°\theta = 60\degree is launching angle and g=9.8m/s2g = 9.8m/s^2 is the gravitational acceleration. Thus, obtain:


hmax=152sin260°29.88.60mh_{max} = \dfrac{15^2\sin^260\degree}{2\cdot 9.8} \approx 8.60m

2. The time of flight is:


T=2vsinθg=215sin60°9.82.65sT = \dfrac{2v\sin\theta}{g} = \dfrac{2\cdot 15\sin60\degree}{9.8} \approx 2.65 s

3. The horizontal range:


R=v2sin2θg=152sin(260°)9.819.9mR = \dfrac{v^2\sin2\theta}{g} = \dfrac{15^2\sin(2\cdot 60\degree)}{9.8} \approx 19.9m

Answer. i) 8.60m, ii) 2.65s, iii) 19.88m.


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