Asked by sydney
A person has the ability to jump with an initial speed of 6.0 m/s at an angle of 35 degree with the ground.
a. How long will the person be in the air?
b. How far horizontally will the person travel?
a. How long will the person be in the air?
b. How far horizontally will the person travel?
Answers
Answered by
Henry
Vo = 6 m/s[35o] = Initial velocity.
Xo = 6*cos35 = 4.91 m/s.=Hor. component of initial velocity.
Yo = 6*sin35 = 3.44 m/s=Ver. component of initial velocity.
a. Y = Yo + g*t = 0
3.44 - 9.8t = 0
9.8t = 3.44
Tr = 3.44/9.8 = 0.3512 s. = Rise time.
Tf = Tr = 0.3512 m/s = Fall time.
T = Tr + Tf = 0.3512 + 0.3512 = 0.702 s.
= Time in air.
b. D=Xo * T = 4.91m/s * 0.702s=3.45 m.
Xo = 6*cos35 = 4.91 m/s.=Hor. component of initial velocity.
Yo = 6*sin35 = 3.44 m/s=Ver. component of initial velocity.
a. Y = Yo + g*t = 0
3.44 - 9.8t = 0
9.8t = 3.44
Tr = 3.44/9.8 = 0.3512 s. = Rise time.
Tf = Tr = 0.3512 m/s = Fall time.
T = Tr + Tf = 0.3512 + 0.3512 = 0.702 s.
= Time in air.
b. D=Xo * T = 4.91m/s * 0.702s=3.45 m.
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