A ball is thrown upwards from the foot of a tower. The ball crosses the top of tower twice after an interval of 4 seconds and the ball reaches ground after 8 seconds, then the height of tower in meters is 10x. Then the value of x is .
$(g = 10 m / s^{2})$

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Solution

$h = u t - \frac{1}{2} g t^{2}$

or $g t^{2} - 2 u t + 2 h = 0$

$t_{1} t_{2} = \frac{2 h}{g}$ and $t_{1} + t_{2} = \frac{2 u}{g} = T$

$∴ (t_{2} - t_{1})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2}$

$16 = 64 - 4 \times \frac{2 h}{g} \Rightarrow h = 60$

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A ball is thrown upwards from the foot of a tower. The ball crosses the top of tower twice after an interval of 4 seconds and the ball reaches ground after 8 seconds, then the height of tower in meters is 10x. Then the value of x is . (g=10m/s2)

h=ut−12gt2 or gt2−2ut+2h=0 t1t2=2hg and t1+t2=2ug=T ∴(t2−t1)2=(t1+t2)2−4t1t2 16=64−4×2hg⇒h=60

A ball is thrown upwards from the foot of a tower. The ball crosses the top of tower twice after an interval of 4 seconds and the ball reaches ground after 8 seconds, then the height of tower in meters is 10x. Then the value of x is .
$(g = 10 m / s^{2})$

$h = u t - \frac{1}{2} g t^{2}$

or $g t^{2} - 2 u t + 2 h = 0$

$t_{1} t_{2} = \frac{2 h}{g}$ and $t_{1} + t_{2} = \frac{2 u}{g} = T$

$∴ (t_{2} - t_{1})^{2} = (t_{1} + t_{2})^{2} - 4 t_{1} t_{2}$

$16 = 64 - 4 \times \frac{2 h}{g} \Rightarrow h = 60$