Gravity Activated

Simply we can apply the law of maximum height which is known as $D=\frac{V^{2}_{. }}{2a}$ where $D$ is the required distance, $V_{. }$ is the initial velocity and $a$ is the acceleration of the body. So the maximum height is $D=\frac{10^2}{2×10}=\boxed {5}$ $\text{Note}:\text{we apply this law when}\space \color{#D61F06}{a}\space \text{regular acceleration}$

Here we can use the equation of motion:

$v^{2}=u^{2}+2as$

Where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration due to gravity, and $s$ is distance travelled. Taking the upwards direction as positive, we know that $a=-10ms^{-2}$ and $u=10ms^{-1}$ . When the ball is at is maximum height, this means it must have stopped moving, therefore $v=0ms^{-1}$ . We can find the value of $s$ by substituting these values into the equation:

$0^{2}=10^{2}+(2 \times (-10) \times s)$

$0=100-20s$

$20s=100$

$s=\boxed{5}$

We could solve this using energy conservation laws.

For this question: Mechanical Energy (ME) = Gravitational Potential Energy (GPE) + Kinetic Energy (KE)

GPE = mgh where m is mass of object, g is acceleration due to gravity and h is change in height. To clarify, this formula works as we assume gravitational field is uniform for such small distance on the surface of the Earth and we take the ground as the zero reference point for the height.

KE=o.5mv^2 where v is the velocity of the object.

All units are SI so they are not written in the algebra that follows:

On the ground, substituting h=0 and v=10:

ME = mg*0 + 0.5m10^2

ME = 50m ... (1)

At maximum height, substituting h=H (max height) and v=0 (instantaneous at the moment)

ME = mgH +0.5m*0

ME = mgH ... (2)

Now since ME is conserved in this situation (there is no energy input or output, only KE is converted to GPE as it goes up) we can equate (1) and (2)

mgH = 50m

H = 50/g

substitutiing g=10

H = 5 (metres)

Please excuse the poor formatting.

Torricelli stated:

v^2 = u^2 +2as

v = final velocity (at peak height = 0m/s)

u = initial speed of 10m/s

a = gravity -10m/s/s

s = distance (height in our case)

Basic algebra yields 5m

For maximum height, we can take final velocity of the ball =0m/s, Apply third equation of motion, 2gh=Vf^2-Vi^2 2(-10)h=(0)^2-(10)^2 -20h=0-100 -20h=-100 h=5m

v=u-g*t, g=10 m/s^2 , u=10 m/s.

v=10-10t
maximum height when v=0, 10-(10)t=0 , t=1 sec
S=ut-(1/2)gt^2
S=(10)(1)-(1/2)(10)(1^2)=5 m. Ans.