Answer:
[tex]R_{horizontal}=2.53\ m[/tex]
Explanation:
Given,
Mass of the steel = 19 g
height reached, h = 1.35 m
angle, θ = 35°
when the ball is projected vertically with initial velocity v.
Maximum height reached is equal to,[tex]h = \dfrac{v^2}{2g}[/tex]
initial velocity,
[tex]v=\sqrt{2gh}[/tex]
[tex]v=\sqrt{2\times 9.8\times 1.35}[/tex]
v = 5.14 m/s
Hence , the maximum velocity is equal to 5.14 m/s
For projectile motion horizontal distance is given by
[tex]R_{horizontal}=\dfrac{v^2sin 2\theta}{g}[/tex]
[tex]R_{horizontal}=\dfrac{5.14^2sin (2\times 35^{\circ}}{9.8}[/tex]
[tex]R_{horizontal}=2.53\ m[/tex]
the horizontal distance travel by the ball is equal to [tex]R_{horizontal}=2.53\ m[/tex]
