Answer:
a) [tex]t=3.199 seconds[/tex]
b) [tex]h = 11.97 m[/tex]
Explanation:
Since this problem belongs to the concept of projectile motion
a) we know,
[tex]Vcos\theta=\frac{R}{t}[/tex]
Where,
V = initial speed
Θ = angle with the horizontal
R = horizontal range
t = Time taken to cover the range 'R'
Given:
V = 27m/s
R = 60m
Θ = 46°
thus,
the equation becomes
[tex]27\times cos46^o=\frac{60}{t}[/tex]
or
[tex]t=\frac{60}{27\times cos46^o}[/tex]
[tex]t=3.199 seconds[/tex]
b)The formula for height is given as:
[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]
where,
g = acceleration due to gravity = 9.8m/s²
substituting the values in the above equation we get
[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]
or
[tex]h = 62.124-50.14[/tex]
or
[tex]h = 11.97 m[/tex]