Answer:
a)t = 1.61 sec.
b)h = 12. 9 m
c)The time required to return on the initial position (T)= 3.22 sec
d)This time is 2 times of the time calculated in part a.
e)The horizontal distance = 58. 64 m
Explanation:
Given that
Upward velocity component(Vo) = 16.1 m/s
Horizontal velocity component(Uo) = 18.2 m/s
a)
When the upward component will become zero then this will be on the highest point
We know that
V= Vo - g.t
0= 16.1 - 10 x t
t = 1.61 sec.
b)
[tex]h=V_o.t-\dfrac{1}{2}gt^2[/tex]
[tex]h=16.1\times 1.61-\dfrac{1}{2}\times 10\times 1.61^2[/tex]
h = 12. 9 m
c)
The time required to return on the initial position = 2 x t
The time required to return on the initial position(T) = 2 x 1.61 = 3.22 sec
The time required to return on the initial position (T)= 3.22 sec
d)
This time is 2 times of the time calculated in part a.
e)
Range R = Uo .T
R = 18.2 x 3.22
R = 58.64 m
The horizontal distance = 58. 64 m
