It depends on both of them.
In fact, the projectile begins its motion with an initial velocity of

and an angle of

. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by

and as it can be seen, this depends on both initial velocity and angle.
Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
Hope this helps!
Answer:
241.8 N.
Explanation:
The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.
Centripetal force = mv^2/r
F = mg + mv²/r
F = m(g + v²/r)
where,
m = mass
= 9 kg
g = acceleration due to gravity
= 9.8 m/s²
v = 3.2 m/s
r = 0.60 m
F = 9 * (9.8 + 3.2²/0.60)
= 241.8 N.
Answer:
a) 17.8 m/s
b) 28.3 m
Explanation:
Given:
angle A = 53.0°
sinA = 0.8
cosA = 0.6
width of the river,d = 40.0 m,
the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,
The river itself was 100 m below the ramp H = 100 m,
(a) find speed v
vertical displacement

putting values h=15 m, v=0.8
............. (1)
horizontal displacement d = vcosA×t = 0.6×v ×t
so v×t = d/0.6 = 40/0.6
plug it into (1) and get

solving for t we get
t = 3.734 s
also, v = (40/0.6)/t = 40/(0.6×3.734) = 17.8 m/s
(b) If his speed was only half the value found in (a), where did he land?
v = 17.8/2 = 8.9 m/s
vertical displacement = 
⇒ 
t = 5.30 s
then
d =v×cosA×t = 8.9×0.6×5.30= 28.3 m