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At maximum height, velocity v = 0 m/s.

March 21, 2026Physics

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2 parts

\sqrt{10} \ \mathrm{s}

10\sqrt{10} \ \mathrm{m/s}

3 steps

Time taken to reach the maximum height

At maximum height, velocity $v = 0$ m/s.
Equation of motion:
$v = u - g t$
Substitute $v = 0$ :
$0 = u - g t$ $u = g t \tag{1}$
Displacement $h = 50$ m.
Equation:
$h = u t - \frac{1}{2} g t^{2}$
Substitute $u = g t$ from (1):
$50 = (g t) t - \frac{1}{2} g t^{2}$ $50 = g t^{2} - \frac{1}{2} g t^{2}$ $50 = \frac{1}{2} g t^{2}$
Substitute $g = 10$ m/s²
$50 = \frac{1}{2} \times 10 \times t^{2}$ $50 = 5 t^{2}$ $t^{2} = 10$ $t = \sqrt{10} \ \mathrm{s}$
\sqrt{10} \ s

Speed of the throw

From equation (1): $u = g t$
Substitute $g = 10$ m/s² and $t = \sqrt{10}$ s:

u = 10 \times \sqrt{10} = 10\sqrt{10} \ \mathrm{m/s}

Alternatively, using $v^{2} = u^{2} - 2 g h$ :

0 = u^{2} - 2 \times 10 \times 50

u^{2} = 1000

u = \sqrt{1000} = \sqrt{100 \times 10} = 10\sqrt{10} \ \mathrm{m/s}

10\sqrt{10} \ m/s

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