Asked by zianna
a projectile is thrown with the speed of 25 meter per seconds in a direction 30° above the horizontal.
-how high does it go.
-how long does it take to reach the maximum height.
-how long does it reach the Projectile along in the air.
-how far does it go.
-how high does it go.
-how long does it take to reach the maximum height.
-how long does it reach the Projectile along in the air.
-how far does it go.
Answers
Answered by
Henry
a. Vo = 25 m/s @ 30 Deg.
Xo = 25*cos30 = 21.7 m/s.
Yo = 25*sin30 = 12.5 m/s.
h = (Y^2-Yo^2)/2g.
h = (0-(12.5)^2) / -19.6=7.97 m.=max ht.
b. Tr = (Y-Yo)/g.
Tr = (0-12.5) / -9.8 = 1.28 s. =Rise time or time to reach max. ht.
c. Tf = Tr = 1.28 s. = Fall time.
Tr+Tf = 1.28 + 1.28 = 2.56 s. = Time in
air.
d. R = Xo*(Tr+Tf).
R = 21.7 * 2.56 = 55.6 m.
Xo = 25*cos30 = 21.7 m/s.
Yo = 25*sin30 = 12.5 m/s.
h = (Y^2-Yo^2)/2g.
h = (0-(12.5)^2) / -19.6=7.97 m.=max ht.
b. Tr = (Y-Yo)/g.
Tr = (0-12.5) / -9.8 = 1.28 s. =Rise time or time to reach max. ht.
c. Tf = Tr = 1.28 s. = Fall time.
Tr+Tf = 1.28 + 1.28 = 2.56 s. = Time in
air.
d. R = Xo*(Tr+Tf).
R = 21.7 * 2.56 = 55.6 m.
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