Answer:
(a) 29.0m/s
(b) 42.05m
Explanation:
(a) Since the motion is in one dimension (i.e vertical), we can use one of the equations of motion as follows;
v = u + at ------------(i)
Where;
v = final velocity of the baseball at a time t
u = initial velocity of the baseball at launch time
a = acceleration due to gravity
t = time taken to reach a certain height
Now;
At maximum height;
v = 0 (velocity is zero when the baseball reaches maximum height)
t = 2.90s
a = -g = -10m/s² (negative sign because the base ball moves upwards against gravity)
Substitute these values into equation (i) as follows;
0 = u - 10(2.90)
u = 29.0m/s
Therefore, the initial velocity of the baseball is 29.0m/s
(b) To get the height reached we use another equation of motion as follows;
Using one of the equations of motion as follows;
v² = u² + 2as -----------------(ii)
Where;
v = final velocity of the baseball
u = initial velocity of the baseball
a = acceleration of the baseball
s = vertical distance covered by the baseball
Remember that;
At maximum height,
v = 0
Also,
u = 29.0m/s (as calculated above)
a = -10m/s²
Substitute these values into equation(ii) as follows;
0² = 29.0² + 2(-10)s
0 = 841 - 20s
841 = 20s
s = [tex]\frac{841}{20}[/tex]
s = 42.05m
Therefore, the maximum height reached is 42.05m
