Answer:
51.96 m/s^-1
Explanation:
a) see the attachment
b) As we know the velocity of the projectile has two component, horizontal velocity v_ox. and vertical velocity v_oy as shown in the figure. At the highest point of the trajectory, the projectile has only horizontal velocity and vertical velocity is zero. Therefore at the highest point of the trajectory, the velocity of the projectile will be
v_ox=v_o*cosФ
=60*cos (30)
= 51.96 m/s^-1
Answer:
(a) a = 2.44 m/s²
(b) s = 63.24 m
Explanation:
(a)
We will use the second equation of motion here:
where,
s = distance covered = 47 m
vi = initial speed = 0 m/s
t = time taken = 6.2 s
a = acceleration = ?
Therefore,
<u>a = 2.44 m/s²</u>
<u></u>
(b)
Now, we will again use the second equation of motion for the complete length of the inclined plane:
where,
s = distance covered = ?
vi = initial speed = 0 m/s
t = time taken = 7.2 s
a = acceleration = 2.44 m/s²
Therefore,
<u>s = 63.24 m</u>