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2 years ago

a stomp rocket takes 1.5 seconds to reach its maximum height what was the initial velocity and what was the maximum height ?

2 answers:

5 0

1) Vf = Vo - gt; Vf = 0 => Vo = gt = 9.8m/s^2 * 1.5s = 14.7 m/s

2) d = Vo*t - gt^2 /2 = 14.7m/s*1.5 - 9.8m/s^2 * (1.5s)^2 / 2 = 11.02 m

Roman55 [17]2 years ago

5 0

Below is the solution:

1) Let the acceleration is a m/s^2 and the velocity after 4 sec is v m/s
=>By v = u + at 
=>v = 4a ------------(i) 
For 1st phase:- 
=>By s = ut + 1/2at^2 
=>s1 = 0 + 1/2 x a (4)^2 
=>s1 = 8a --------------(ii) 
For 2nd phase:- 
=>By s = vt 
=>s2 = 4a x (9.1-4) 
=>s2 = 20.4a -------------(iii) 
=>By (ii) + (iii):- 
=>s1+s2 = 8a + 20.4a 
=>100 = 28.4a 
=>a = 3.52 m/s^2 
2)(a) By v = u -gt 
=>0 = u - 9.8 x 1.5 
=>u = 14.7 m/s 
(b) By v^2 = u^2 - 2gh 
=>0 = (14.7)^2 - 2 x 9.8 x h 
=>h = 11.03 m

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