Question

if an object is thrown vertically upwards with speed 49m-1. how long does it take to complete upward journey? what maximum height does it achieve

Answer 1

Hey.

$inital \: velocity \: u \: = 49 \: m {s}^{ - 1} \\ \\ as \: thrown \: upward \: so \: \: a = g = 9.8m {s}^{ - 2} \\ \\ velocity \: at \: maximum \: height \: v = 0 \: m {s}^{ - 1} \\ \\ as \: we \: know \: \: v \: = u \: + \: at \\ \\ so \: \: 0 = 49 + ( - 9.8) \times t \\ \\ so \: \: t \: = \frac{ - 49}{ - 9.8} \\ \\ so \: \: t = 5 \: s \: \\ \\ also \: \\ \\ {v}^{2} - {u}^{2} = 2as \\ \\ so \: \: {0}^{2} - {49}^{2} = 2 \times ( - 9.8 )\times h \\ \\ or \: \: h \: = \frac{ {49}^{2} }{2 \times 9.8} \\ \\ so \: \: h \: = 122.5 \: m$

So, time taken to complete upward journey will be 5 seconds and the maximum height gained is 122.5 m.

Thanks.

Answer 2

Answer:

Hi mate

here is the answer

I did it on copy

thanks ❣