Recent content by loku

so is the height 100m ??

Well in that case the initial displacement will be 5 m. so by using the equation, you gave y(t) = 5 + v*5 + (1/2)(-10)t^2 for t= 5sec and v=5m/s y(5) = 5+ 5*5 +(1/2)(-10)*25 y(5) = -95m but I know that's not correct

I haven't learned about 2d motion. So simply, can we use the equation s=ut+(1/2)at^2 please?

shouldn't the displacement of the body be the height of the tower, since it was thrown from the height of the tower?

s = 5 + 5*5 - 0.5*10*25 s = 5 + 25 -125 s = - 95 m

but still then by using the equation you gave, we don't get the correct answer.

we are taught: y(t) = vy0∗t+1/2 ayt^2

I am not sure what s0 term is missing. I know only the three equations of motion.

u=5m/s t=5sec the height will be the displacement s=ut - (1/2)gt^2 s=5*5-0.5*10* 25 s=-100m But the answer is 273.6m