Answer:
99
Explanation:
First strike=25 in
Second strike=4/5*25=20 in
Third strike=4/5*20=16 in
Fourth strike=4/5*16=12.8 in
Fifth strike=4/5*12.8=10.24 in
sixth strike=4/5*10.24=8.192 in
seventh strike=4/5*8.192=6.5536 in
Total=(25+20+16+12.8+10.24+8.192+6.5536) in=98.7856 in
Taking this as a geometric series
where a is initial value taken as 25 for this case and r is rate taken as 4/5 or 0.8 in our case
To the nearest inches, sum is 99 in
Answer:
The maximum height reached by the rocket is 486.53 m
Explanation:
Given;
initial velocity of the rocket, u = 0
acceleration of the rocket, a= 20 m/s²
duration of the rocket first motion, t = 4 s
The distance traveled by the rocket before its thrust failed
h₁ = ut + ¹/₂at²
h₁ = 0 + ¹/₂ x 20 x 4²
h₁ = 160 m
The second distance moved by the rocket is calculated as follows;
The velocity of the rocket before its thrust failed;
v = u + at
v = 0 + 20 x 4
v = 80 m/s
This becomes the initial velocity for the second stage
At maximum height, the final velocity = 0
The maximum height reached by the rocket = h₁ + h₂
= 160 + 326.53
= 486.53 m