In 1940, Emanuel Zacchini flew about 53m as a human cannonball, a record that re
ID: 1698287 • Letter: I
Question
In 1940, Emanuel Zacchini flew about 53m as a human cannonball, a record that remains unbroken. His initial velocity was 24.2 m/s at an angle ?. Find ? and the maximum height "h" Emanuel achieved during the flight. Ignore the effects of air resistance.The answer key says ?=31.3? and h=8.06 m
In the link, I don't understand how they got R=(vo^2)(sin^2 ?)/(g)
I've gotten R=vocos?t
and then t=53/vocos?
How did they come up with the equation in the link?
Explanation / Answer
R=vo^2*sin(2*theta)/g:--->9.8*53/24.2^2=0.887--->arcsin(0.887)/2=31.2 degrees R=vo^2*sin(2*theta)/g=53 meters--->voy=vo*sin(theta)=24.3*sin(31.2 degrees)=12.552 m/s--->vox=vo*cos(theta)=24.2*cos(31.2 degrees)=20.69 m/s--->ay=9.8 m/s^2--->ax=0 max height:voy^2/(2*g)=12.552^2/(2*9.8)=8.04 m max ht t:voy/g=12.552/9.8=1.281 s--->total flight time=1.281*2=2.562 s degrees:31.2 height in meters:8.04 ht time in s:1.281 total flight time in s:2.562