Answer:
Explanation:
given,
speed of the pebble = 14 m/s
height of the building = 31 m
ignoring air resistance
a) horizontally
V_x = 14 m/s
time = [tex]\sqrt{\dfrac{2h}{g}}[/tex]
time = [tex]\sqrt{\dfrac{2\times 31}{9.8}}[/tex]
times = 2.52 s
v_y = 9.8 x 2.52
v_y = 24.7 m/s
resultant velocity
= [tex]\sqrt{24.7^2+14^2}[/tex]
= 28.39 m/s
b) vertically straight up
max height = [tex]\dfrac{v^2}{2g}[/tex]
= [tex]\dfrac{15^2}{2\times 9.8}[/tex]
= 11.47 m
height covered = 31 + 11.47
= 42.47 m
velocity = [tex]\sqrt{2gh}[/tex]
= [tex]\sqrt{2\times 9.8 \times 42.47}[/tex]
= 28.85 m/s
c) vertically down
v² = u² + 2as
v² = 2 × 9.8 × 31 +14²
v² = 803.6
v = 28.35 m/s
(a) The final velocity of the pebble when it is fired horizontally is 14 m/s.
(b) The final velocity of the projectile when fired vertically straight up is zero.
(c) The final velocity of the pebble when it is fired vertically straight down is 28.31 m/s.
The time of motion of the pebble is calculated as follows;
[tex]h = v_0t + \frac{1}{2} gt^2\\\\31 = 14t + 4.9t^2\\\\4.9t^2 + 14t - 31 = 0\\\\t = 1.46 \ s[/tex]
The final velocity of the pebble when it is fired horizontally is calculated as follows;
[tex]v_f = v_i\\\\v_f = 14 \ m/s[/tex]
The final velocity of the projectile when fired vertically straight up is zero, because velocity is zero at maximum height.
The final velocity of the pebble when it is fired vertically straight down is calculated as follows;
[tex]v_f = v_i + gt\\\\v_f = 14 + 9.8(1.46)\\\\v_f = 28.31 \ m/s[/tex]
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