using the eqaution of linear motion we have
X = Ux*t
D= v cos 37*t
1 = v cos 37 *t
t = 1/v cos 37 ---------(1)
Y = Uyt - 0.5gt^2
using the value of t from (1) we have
-0.3 = v sin 37* ( 1/v cos 37) - 0.5*9.8*(1/v cos 37)^2
-0.3 = tan 37 - 4.9 / (v cos 37)^2
4.9 / (v cos 37)^2 = 0.3 + tan 37
v = 2.7 m/s
so the initial speed of the frog is 2.7 m/s
A frog jumps from a rock through a distance of 1m (horizontally) to the center of...