h = 122.5 m g = 9.8 m / s 2 h=122.5 \;m \\
g = 9.8 \;m/s^2 h = 122.5 m g = 9.8 m / s 2
a) t= 2 sec
Distance traveled in 2 sec
S = 1 2 g t 2 S = 1 2 × 9.8 × 2 2 = 19.6 m v 2 − u 2 = 2 g S v 2 − 0 2 = 2 × 9.8 × 19.6 v = 384.16 = 19.6 m / s S = \frac{1}{2}gt^2 \\
S = \frac{1}{2} \times 9.8 \times 2^2 = 19.6 \;m \\
v^2 -u^2 = 2gS \\
v^2 -0^2 = 2 \times 9.8 \times 19.6 \\
v = \sqrt{384.16} = 19.6 \;m/s S = 2 1 g t 2 S = 2 1 × 9.8 × 2 2 = 19.6 m v 2 − u 2 = 2 g S v 2 − 0 2 = 2 × 9.8 × 19.6 v = 384.16 = 19.6 m / s
b) Using the equation of motion
h = u t + 1 2 g t 2 122.5 = 0 × t + 1 2 × 9.8 × t 2 t 2 = 2 × 122.5 9.8 t = 25 = 5 s e c h=ut+ \frac{1}{2}gt^2 \\
122.5 = 0 \times t + \frac{1}{2} \times 9.8 \times t^2 \\
t^2 = \frac{2 \times 122.5}{9.8} \\
t = \sqrt{25} = 5 \;sec h = u t + 2 1 g t 2 122.5 = 0 × t + 2 1 × 9.8 × t 2 t 2 = 9.8 2 × 122.5 t = 25 = 5 sec
c)
v 2 = u 2 + 2 g h v 2 = 0 2 + 2 g h v = 2 g h v = 2 × 9.8 × 122.5 v = 49 m / s v^2=u^2 +2gh \\
v^2 = 0^2 +2gh \\
v= \sqrt{2gh} \\
v= \sqrt{2 \times 9.8 \times 122.5} \\
v = 49 \;m/s v 2 = u 2 + 2 g h v 2 = 0 2 + 2 g h v = 2 g h v = 2 × 9.8 × 122.5 v = 49 m / s
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