Hello, could you guys let me know if this looks right? Thanks in advance.
A ball is fired at a speed of 8 m/s at 20 degrees above the horizontal from the top of a 5m high wall.
a) what is the total velocity at the maximum height?
b) how far from the launcher will the ball hit the ground?
a)
v(cos(20)) = 8cos(20 deg)
=3.26 m/s
b)
s = ut + 0.5at^2
u = 8*sin20 =
-5 =(8sin(20deg))t - 0.5(9.8)t^2
t = 1.33 s
= 8*cos(20 deg)*1.33
=4.34m
Vo = 8m/s[20o].
Xo = 8*Cos20 = 7.52 m/s.
Yo = 8*sin20 = 2.74 m/s.
a. Y = 0 m/s at max ht.
V = sqrt(Xo^2+Y^2) = sqrt(7.52^2 + 0^2) = 7.52 m/s.
b. Y = Yo + g*Tr.
0 = 2.74 - 9.8Tr, Tr = 0.279 s. = Rise time.
b. h = ho + Yo*Tr + 0.5g*Tr^2.
h = 5 + 2.74*0.279 - 4.9*0.279^2 = 5.38 m. Above gnd.
h = o.5g*Tf^2.
5.38 = 4.9Tf^2, Tf = ? = Fall time.
d = Xo*(Tr+Tf).