97. Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the initial
speed:
v
0
=
x
cos θ
0
g
2 (x tan θ
0
− y)
which yields v
0
= 23 ft/s for g = 32 ft/s
2
, x = 13 ft, y = 3 ft and θ
0
= 55
◦
.
97. Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the initial
speed:
v
0
=
x
cos θ
0
g
2 (x tan θ
0
− y)
which yields v
0
= 23 ft/s for g = 32 ft/s
2
, x = 13 ft, y = 3 ft and θ
0
= 55
◦
.