A ball is projected horizontally from a height of 100m from the ground with a speed of 20m/s .the time taken to reach the ground,will be.take g=10m/s2
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We know that s = ut - (1/2)at²
hence, 100 = 20t - (1/2)10t²
100 = 20t - 5t²
5t² - 20t + 100 = 0
t² - 4t + 20 = 0
now t=2\left(\sqrt{6}-1\right),\:t=-2\left(1+\sqrt{6}\right)
∴ t = 2.89 sec
hence, 100 = 20t - (1/2)10t²
100 = 20t - 5t²
5t² - 20t + 100 = 0
t² - 4t + 20 = 0
now t=2\left(\sqrt{6}-1\right),\:t=-2\left(1+\sqrt{6}\right)
∴ t = 2.89 sec
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