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.
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?
1 answer