A ball is thrown upwards from the ground with an initial speed of $u$ . The ball is at a height of $80 m$ at two times, the time interval being $6 s$ . Find $u$ . Take $g = 10 m / s^{2}$

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Solution

Time taken to go from B to C or C to D = $\frac{6}{2} = 3 s e c$
Speed at D = $0 + 10 \times 3 = 30 m / s e c$
As speed a tA = Speed at E = $u$
We use $v^{2} = u^{2} + 2 a s$
$u^{2} = 30^{2} + 2 \times 10 \times 80$
$= 900 + 1600 = 2500$
$∴ u = 50 m / s e c$

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A ball is thrown upwards from the ground with an initial speed of u. The ball is at a height of 80m at two times, the time interval being 6s. Find u. Take g=10m/s2

Time taken to go from B to C or C to D = 62=3secSpeed at D =0+10×3=30m/secAs speed a tA = Speed at E = uWe use v2=u2+2asu2=302+2×10×80=900+1600=2500 ∴u=50m/sec

A ball is thrown upwards from the ground with an initial speed of $u$ . The ball is at a height of $80 m$ at two times, the time interval being $6 s$ . Find $u$ . Take $g = 10 m / s^{2}$

Time taken to go from B to C or C to D = $\frac{6}{2} = 3 s e c$
Speed at D = $0 + 10 \times 3 = 30 m / s e c$
As speed a tA = Speed at E = $u$
We use $v^{2} = u^{2} + 2 a s$
$u^{2} = 30^{2} + 2 \times 10 \times 80$
$= 900 + 1600 = 2500$
$∴ u = 50 m / s e c$