To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to



The vertical component of velocity is

Here,
h = Height
g = Gravitational acceleration
t = Time
= Vertical component of velocity



The direction of the velocity will be given by the tangent of the components, then



The magnitude is given vectorially as,



Therefore the angle is 55.59° and the velocity is 26.37m/s
According to the position vs time graph, the <em>average</em> <em>velocity</em> of the motorcycle is the change in position divided by the change in time. Also, note that the slope is linear and positive throughout the 5 hours, it doesn't change direction.
Therefore, we have
Avg velocity = change in direction/change in time
Avg velocity = (150km - 30km)/(5h - 0h)
Avg velocity = 24km/hr south.
Answer:
v0 = 25 m/s
vf = 0 m/s
a = -9.80 m/s^2
change in x = 31.89m
but that's only 1/3 of the hight, so i time it by 3 to get 96m