Answer:
Explanation:
given,
initial speed of the shot = 12.0 m/s
angle = 40°
height at which shot leaves her hand = 1.80 m
v_x = 12 cos 40° = 9.19 m/s
v_y = 12 sin 40° = 7.71 m/s
time to reach maximum height =
= [tex]\dfrac{v_y}{g}[/tex]
= [tex]\dfrac{7.71}{9.8}[/tex]
= 0.787 s
[tex]h = v_y t + \dfrac{1}{2}gt^2[/tex]
h = 7.71 × 0.787 - 0.5 × 9.81 × 0.787²
h = 3.03 m
the maximum height attain = 3.03 + 1.8 = 4.83 m
now free fall from the maximum height
[tex]h =\dfrac{1}{2}gt^2[/tex]
[tex]4.83 = \dfrac{1}{2}\times 9.8 \times t^2[/tex]
t = 0.9928 s
total time = 0.9928 + 0.787 = 1.7798 s
range =
d = vₓ t
d = 16.36 m
