Answer:
(a) 20 m/s (j) m/s
(b) 20√3 m/s (i) m/s
(c) 2.04 s
(d) 20.4 m
Explanation:
In order to solve the problem, you have to apply the <em>Projectil Motion</em> equations.
For part (a) and (b) you have to obtain the components of the initial velocity vector. The direction forms a 30° angle to the horizontal and the modulus (speed) was given. Therefore:
Applying trigonometric identities (Because the initial velocity is the hypotenuse of a right triangle with angle 30° to the horizontal)
Vx: 40Cos(30°)=20√3 m/s
Vy:40Sin(30°)= 20 m/s
The initial velocity in the y direction is: 20 m/s (j) m/s
The initial velocity in the x direction is: 20√3 m/s (i) m/s
Where i and j are the unit vectors.
For part (c) you have to apply the following vertical motion equation:
Vy=Voy-gt
where Voy is the initial velocity, g is gravity and t is the time
The ball reaches its max height when Vy=0 therefore:
0=Voy-gt
Solving for t:
t=Voy/g=20/9.8= 2.04 seconds
For part (d) you have to apply the other vertical motion equation which is:
y=yo+Voyt-0.5gt²
Where yo is the initial position.
Replacing t=2.04 s, yo=0 m, Voy=20 m/s and solving for y:
y=0+(20)(2.04)-(0.5)(9.8)(2.04)²
y=20.4 m
