Respuesta :
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 Projectil Motion 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
