Answer to Question #166001 in Astronomy | Astrophysics for wadzani

Question #166001
a stone propelled from a catapault with a speed of 50m/s attains a height
of 100 m. calculate;the time of flight,the angle of projection,the range attained
1
Expert's answer
2021-02-24T12:49:51-0500


a) Time of flight for a projectile, T=2UsinθgT=2Usin\frac{\theta}{g}

U=50m/sU = 50 m/s

θ=?\theta = ?

g=10m/s2g = 10 m/s^2

However, maximum height, H=U2sin2H = U^2sin^2θ2g\frac{\theta}{2g}

H=100mH = 100 m

Hence, 100=502sin2θ2×10100 = 50^2 sin^2\frac{\theta}{2\times10} ​

sin2θ=2×10×100502=0.8sin^2\theta = \frac{2\times10\times100}{50^2}= 0.8

sinθ=0.8944sin\theta = 0.8944

Therefore, T=2×50×0.894410=8.9sT = \frac{2\times50\times0.8944}{10} = 8.9 s

b) sinθ=0.8944sin\theta = 0.8944

Using Natural Sine Table (Trigonometrical Tables)

θ\theta, angle of projection = 63,4o1'

c) Range, R=U2sin2θg=502sin(2×63.4o1)10=250×0.8007=200.2mR = \frac{U^2sin2θ}{g} = \frac{50^2sin(2\times63.4^o1')}{10} = 250 \times 0.8007 = 200.2 m

Answer: a) T = 8.9 s; b) θ\theta = 63,4o1'; c) R = 200.2 m.


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