Asked by Justin
A ball is fired from a height of 1.4 m above ground, at a speed of 3.6 m/s, and at an angle of 25° above horizontal. (a) What is the peak height of the ball above ground? (b) Where does the ball land? Ignore Earth's atmosphere, curvature, and rotation.
This is what I got:
a)
(⌊3.6⌉^2 〖sin〗^2(25))/2(9.80) =1.18m+1.4m= 2.58m
b)
(25)(3.6)=90m
Is this right?
This is what I got:
a)
(⌊3.6⌉^2 〖sin〗^2(25))/2(9.80) =1.18m+1.4m= 2.58m
b)
(25)(3.6)=90m
Is this right?
Answers
Answered by
henry2,
Vo = 3.6m/s[25o].
Xo = 3.6*Cos25 = 3.26 m/s. = Hor. component of Vo.
Yo = 3.6*sin25 = 1.52 m/s. = Ver. component of Vo.
a. Y^2 = Yo^2 + 2g*h = 0,
1.52^2 + (-19.6)h = 0,
h = 0.118 m. above launching point.
1.4 + 0.118 = 1.52 m. above gnd.
b. Y = Yo + g*Tr = 0,
1.52 + (-9.8)Tr = 0,
Tr = 0.155 s. = Rise time.
h = 0.5g*Tf^2 = 1.52,
4.9Tf^2 = 1.52,
Tf = 0.557 s. = Fall time.
d = Xo * (Tr+Tf) = 3.26 * (0.155+0.557) meters.
Xo = 3.6*Cos25 = 3.26 m/s. = Hor. component of Vo.
Yo = 3.6*sin25 = 1.52 m/s. = Ver. component of Vo.
a. Y^2 = Yo^2 + 2g*h = 0,
1.52^2 + (-19.6)h = 0,
h = 0.118 m. above launching point.
1.4 + 0.118 = 1.52 m. above gnd.
b. Y = Yo + g*Tr = 0,
1.52 + (-9.8)Tr = 0,
Tr = 0.155 s. = Rise time.
h = 0.5g*Tf^2 = 1.52,
4.9Tf^2 = 1.52,
Tf = 0.557 s. = Fall time.
d = Xo * (Tr+Tf) = 3.26 * (0.155+0.557) meters.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.