1. A model rocket leaves the ground, heading straight up at45 m/s. (a) What is i

Question

1. A model rocket leaves the ground, heading straight up at45 m/s.

(a) What is its maximum altitude?
1 m
What are its speed and altitude at each of the following times?
(b) 1 s
2 m/s
3 m
(c) 3 s
4 m/s
5 m
(d) 6 s
6 m/s
7 m

2. A Frisbee is lodged in a tree branch, 6.4 m above the ground. A rock thrown from below mustbe going at least 3 m/s to dislodge theFrisbee. How fast much such a rock be thrown upward, if it leavesthe thrower's hand 1.1 m above theground?
1 m/s
(a) What is its maximum altitude?
1 m
What are its speed and altitude at each of the following times?
(b) 1 s
2 m/s
3 m
(c) 3 s
4 m/s
5 m
(d) 6 s
6 m/s
7 m

Explanation / Answer

a.Applying law of conservation of energy we have, KE=PE 1/2*mv2=mgh h=v2/2g h=452/2g h=103.3163m b. For 1s we have, v=u-gt v=45-9.8*1 v=35.2m/s v2-u2=2*-g*h h=(35.22-452)/2-g h=40.1m Similarly for others. The height from 1.1 is h=6.4-1.1=5.3m Applying law of conservation of energy we have as, PE=KE mgh=1/2*mv2 v=2gh v=2*9.8*5.3 v=10.1921m/s Hence we get by it. v=2gh v=2*9.8*5.3 v=10.1921m/s Hence we get by it.

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