Question

State whether this statement is true or false. An object thrown upwards has a constant acceleration but zero velocity at the maximum height.
(A) True
(B) False

Answer 1

Hint : An object which is thrown upward will act solely under gravity. At the point where a body changes direction due to a force being applied in the opposite direction, it will momentarily be at rest.

Formula used: In this solution we will be using the following formulae;
$W = mg$ , where $W$ is the weight of a body, $m$ is the mass of the body, $g$ is the acceleration due to gravity.
${F_{net}} = ma$ where ${F_{net}}$ is the net force acting on a body, $a$ is the acceleration of the body.

Complete step by step answer:
When a body is projected vertically upward, perhaps thrown with the hand or shot with some machine, the moment the force of the machine or the hand stops acting on the body, the only force which will begin to act is the gravitational pull of the earth on that body. The gravitational force is solely dependent on the weight of the body, and the weight is given by
$W = mg$ , where $W$ is the weight of a body, $m$ is the mass of the body, $g$ is the acceleration due to gravity.
If we perform a Newton’s second law analysis on the body, we have
${F_{net}} = ma$ where ${F_{net}}$ is the net force acting on a body, $a$ is the acceleration of the body.
Hence, since only the weight act on the body, then
$W = mg = ma$
$\Rightarrow g = a$
Since $g$ is constant, then, the body has a constant acceleration.
However, since acceleration comes down and the body is initially going upward, during ascent the downward acceleration reduces the upward velocity to zero momentarily, then increases it downward. Hence, the statement is true.
Thus, the correct answer is A.

Note:
For clarity, we assumed that there is no air resistance acting on the body during the journey. However in reality, air resistance which is proportional to velocity always act on a body moving through the atmosphere, no matter the direction. Hence, the acceleration is not actually constant.