Answer:
19. a) 48.3 m
b) 88.2 m
20. a) t ≅ 6.38 s
b) ≅270.1 m
Explanation:
Use basic linear motion equations for each of these questions, correctly.
s = ut + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex]
a) applying the equation to the horizontal direction,
s = ut = 11.5×4.2 = 48.3 m ( as gravitational acceleration for horizontal direction is zero)
b) applying the equation to the vertical direction in downward direction,
s = h( height ) =0 + [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex] ( as initial velocity in vertical direction is zero.)
h = 0.5×10×[tex]4.2^{2}[/tex]
=88.2 m
20) Using the same equation,
a) s = ut + [tex]\frac{1}{2}[/tex]a[tex]t^{2}[/tex] apply it for vertical motion just as it reachs the ground in upwards direction,
0 = usin37°t - [tex]\frac{1}{2}[/tex]g[tex]t^{2}[/tex]
53(0.6081) = 0.5×10×t
t ≅ 6.38 s
b) again same equation in horizontal direction,
s = ucos37°t = 53×0.7986×6.38
≅270.1 m
