v0 The time tmax it takes for the projectile to reach its maximum height is vo c

Question

v0 The time tmax it takes for the projectile to reach its maximum height is vo cos 1. tmax- 2. tmar - 3. t 4. t vo cos uo u0 sin 2 VO Sin 7. tmax- Vog 9 017 (part 2 of 3) 10.0 points Given vo = 21.8 m/s and = 30°, what is the speed of the projectile when it reaches its maximum height y -ymazi (i.e., at point A in the figure)? Answer in units of m/s. 018 (part 3 of 3) 10.0 points Find the speed of the projectile, on its way down, at the height y in the figure) i.e. at point B Answer in units of m/s.

Explanation / Answer

here,

the angle with horizontal is alpha

initial speed is v0

let the time taken to reach maximum height be t

for vertical direction

v = v0 * sin(alpha) - g * t

t = v0 * sin(alpha)/g

so the correct option is 5) tmax = v0 * sin(alpha)/g

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