Answer:
a)x= 12.50 m,y= 7.84 m
b)x= 19.32 m,y= 7.02 m
c)x= 29.44 m,y= -0.81 m
Explanation:
Given that
Initial velocity U= 17 m/s
θ = 48°
This is the case of projectile motion
So the horizontal component of velocity U = U cosθ
u = U cosθ
u = 17 cos 48°
u=11.37 m/s
The vertical component of velocity U = U sinθ
Vo= U sinθ
Vo= 17 sin 48°
Vo = 12.63 m/s
a)
t= 1.1 s
We know that in projectile motion horizontal component of velocity will remain constant.
So the horizontal distance ,x
x = u .t
x = 11.37 x 1.1 m
x= 12.50 m
In vertical direction
[tex]y=V_ot-\dfrac{1}{2}gt^2[/tex]
[tex]y=12.63\times 1.1-\dfrac{1}{2}\times 10\times (1.1)^2[/tex]
y= 7.84 m
b)
t= 1.7 s
So the horizontal distance ,x
x = u .t
x = 11.37 x 1.7 m
x= 19.32 m
In vertical direction
[tex]y=V_ot-\dfrac{1}{2}gt^2[/tex]
[tex]y=12.63\times 1.7-\dfrac{1}{2}\times 10\times (1.7)^2[/tex]
y= 7.02 m
c)
t= 2.59 s
So the horizontal distance ,x
x = u .t
x = 11.37 x 2.59 m
x= 29.44 m
In vertical direction
[tex]y=V_ot-\dfrac{1}{2}gt^2[/tex]
[tex]y=12.63\times 2.59-\dfrac{1}{2}\times 10\times (2.59)^2[/tex]
y= -0.81 m
