Question

Time is taken to reach the maximum height t ( 1= )
( mathrm{u} / mathrm{g} )
If ( mathrm{t}_{2} ) is the time taken to hit the ground, ( -mathrm{H}=mathrm{ut}_{2}-frac{1}{2} mathrm{gt}_{2}^{2} )
[
t_{2}=n t_{1}
]
( mathrm{i} . mathrm{e},-mathrm{H}=mathrm{u} frac{mathrm{nu}}{mathrm{g}}-frac{1}{2} mathrm{g} frac{mathrm{n}^{2} mathrm{u}^{2}}{mathrm{g}^{2}} )
( -mathrm{H}=frac{mathrm{nu}^{2}}{mathrm{g}}-frac{1}{2} frac{mathrm{n}^{2} mathrm{u}^{2}}{mathrm{g}} )
( 2 g H=n u^{2}(n-2) )
Time is taken to reach the maximum height ( t 1= )
( mathrm{u} / mathrm{g} )
If ( mathrm{t}_{2} ) is the time taken to hit the ground, ( -mathrm{H}=mathrm{ut}_{2}-frac{1}{2} mathrm{gt}_{2}^{2} )
[
t_{2}=n t_{1}
]
( mathrm{i} . mathrm{e},-mathrm{H}=mathrm{u} frac{mathrm{nu}}{mathrm{g}}-frac{1}{2} mathrm{g} frac{mathrm{n}^{2} mathrm{u}^{2}}{mathrm{g}^{2}} )
( -mathrm{H}=frac{mathrm{nu}^{2}}{mathrm{g}}-frac{1}{2} frac{mathrm{n}^{2} mathrm{u}^{2}}{mathrm{g}} )
( 2 g H=n u^{2}(n-2) )

# (c) 4y = 2X - 25 34. From a tower of height H, a particle is thrown vertically upwards with a speed u. The time taken by the particle to hit the ground, is n times that taken by it to reach the highest point of its path. The relation between H, u and n is: (a) 2gH = nu?(n-2) (b) gH = (n - 2)u? (c) 2gH = n²u2 (d) gH = (n - 2)?u?

Solution