Question

Hi sir here with problem if you are free pls provide solution for this



1) An object is thrown upward with a speed of 25ms-1.How high will it be when the speed is 12ms-1 ?
A) S=ut + 1/2 at^2
B) S= (u+v)t/2
C) V=u+ at
D) V^2=U^2+2aS

Which equation will allow the problem to be solved in a single calculation?

Answer 1

An object is thrown upward with a speed of 25ms-1. How high will it be when the speed is 12ms-1?

A) $S = ut + \frac{1}{2} at^2$

B) $S = (u + v)t / 2$

C) $V = u + at$

D) $V^2 = U^2 + 2aS$

Which equation will allow the problem to be solved in a single calculation?

Solution:

The speed at the moment of time $t = 0$ will be $u = 25 \, \text{ms}^{-1}$.

The speed at the height "S" will be $V = 12 \, \text{ms}^{-1}$. We don't know in which moment of time will it be.

A) $S = ut + \frac{1}{2} at^2$

We cannot use this equation, because we don't know the time «t».

B) $S = (u + v)t / 2$

The same situation: we cannot use this equation, because we don't know the time «t»

C) $V = u + at$

The same.

D) $V^2 = U^2 + 2aS$

We know all values in this equation. So we can use it for finding «S».

Answer:

Equation «D» will allow the problem to be solved in a single calculation.