Answer:
(a) The ball's initial speed is 8.4 m/s
(b) The height of the ball is 3.28 m
Explanation:
Given;
maximum height of the ball, H = 3.6 m
Apply conservation of energy;
¹/₂mu² + mgh = ¹/₂mv² + mgH
Where;
m is the mass of the ball
u is the initial velocity of the ball
v is the final velocity of the ball
h is the initial height of the ball
H is the maximum height of the ball
(a) The ball's initial speed, u;
¹/₂mu² + mgh = ¹/₂mv² + mgH
¹/₂u² + gh = ¹/₂v² + gH
make u the subject of the formula
[tex]u = \sqrt{v^2 +2gH -2gh} \\\\[/tex]
at maximum height, the final velocity = 0
maximum height H = 3.6 m
initial height, h = 0
g is acceleration due to gravity = 9.8 m/s²
[tex]u = \sqrt{0^2 +2*9.8*3.6 -2*9.8*0} \\\\u = \sqrt{2*9.8*3.6} \\\\u = 8.4 \ m/s[/tex]
(b) The height of the ball when it has a speed of 2.5 m/s
v = 2.5 m/s
[tex]u = \sqrt{v^2 +2gH -2gh} \\\\u^2 = v^2 +2gH -2gh \ (h = 0)\\\\u^2 = v^2 +2gH\\\\2gH = u^2 - v_2\\\\H = \frac{u^2 - v^2}{2g} \\\\H = \frac{8.4^2 - 2.5^2}{2*9.8}\\\\H = 3.28 \ m[/tex]
A) The ball's initial speed is : 8.4 m/s
B) The height of the ball when it has a speed of 2.5 m/s is : 3.28 m
Given data :
Maximum height reached by ball = 3.60 m
applying conservation of energy relation
¹/₂mu² + mgh = ¹/₂mv² + mgH -------- ( 1 )
where : u = initial velocity , v = final velocity , H = max height , h = initial height
Given that the value of m is not given equation ( 1 ) becomes
¹/₂u² + gh = ¹/₂v² + gH --- ( 2 )
solve for u ( initial velocity )
u = [tex]\sqrt{v^2 +2gH -2gh}[/tex] ----- ( 3 )
where : g = 9.8 m/s², v = 0 , H = 3.60 m , h = 0
insert values into equation ( 3 )
u ( initial speed of ball ) = 8.4 m/s
applying equation ( 3 )
u = [tex]\sqrt{v^2 +2gH -2gh}[/tex]
where : u = 8.4 m/s , v = 2.5 m/s , H = ? , h = 0
solve for H
H = u² - v² / 2g
= ( 8.4² - 2.5² ) / ( 2 * 9.8 )
= 3.28 m
Hence we can conclude that The ball's initial speed is : 8.4 m/s and The height of the ball when it has a speed of 2.5 m/s is : 3.28 m.
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