Question

a body is thrown up with speed of 30m/s. how hight will it go and what is the total time of flight? g=10m/s​

Answer 1

Answer:

When the body is thrown up,the acceleration, a= -g

In the second kinematic equation of motion,s becomes

$s = ut - \frac{1}{2} gt {}^{2} ........................(1)$

Let us take the first kinematic equation of motion to find time of flight.

$v = u - gt \: \: \: \: \: \:since \: a = - g$

When the body reaches up , v = 0

$therefore \: 0 = 30 - 10t \: \: \: \: \: since \: g = 10 \: u = 30(given) \\ \: = > 10t = 30 \\ \: \: \: = > t = 3s$

Now,

$(1) = > h = 30(3) - \frac{1}{2} 10(3) {}^{2} \: \: \: \: \: take \: s \: as \: h$

$\: \: \: = 90 - \frac{1}{2} (90)$

$\: \: \: = 90 - 45$

$\: \: \: = 45m$

Thus height =45m and time of flight= 3s