Question
Question: Velocity of the top most point is 10m/s, if the object is projected at an angle of \[{30^{\circ}}\]....
Velocity of the top most point is 10m/s, if the object is projected at an angle of 30∘. Calculate the maximum height.
Solution
Hint : When an object projected at an angle then motion of the object is known as projectile motion. Projectile motion is always projected in the form of parabola.
The maximum height of an object is calculated as, H = 2gv2sin2θ
Where,
H is the maximum height
v is the velocity of the object
θ is angle of projection
g is the acceleration due to gravity,
Complete step by step answer:
Given that:
Velocity of the top most point is, v = 10sm
Angle of projection is, θ=30∘
Consider that acceleration due to gravity is, g = 10s2m
The maximum height is calculated by the vertical velocity (i.e. velocity in Y-direction) and the acceleration due to gravity. It does not depend on the horizontal velocity (i.e. velocity in X-direction).
The maximum height of an object is calculated as, H = 2gv2sin2θ
Now put all the required values in the above equation,
{\text{H = }}\dfrac{{{{\text{v}}^{\text{2}}}{\text{si}}{{\text{n}}^{\text{2}}}{\theta }}}{{{\text{2g}}}} \\\
= \dfrac{{{{\left( {{\text{10}}} \right)}^{\text{2}}}{\text{si}}{{\text{n}}^{\text{2}}}\left( {{{30}^{\circ}}} \right)}}{{{\text{2}} \times {\text{10}}}} \\\
= \dfrac{{100 \times {{\left( {\dfrac{1}{2}} \right)}^2}}}{{20}} \\\
= \dfrac{{10}}{8} \\\
= 1.25{\text{m}} \\\
Hence, the maximum height of the object is, H = 1.25m
Note:
Some common examples of projectile motion are cricket ball and football. The projectile motion occurs in two steps, the first step is called horizontal motion which has no acceleration and the second step is called vertical motion which has an acceleration in the form of acceleration due to gravity.
In other words we can say that when a body is projected at any angle then the only acceleration acting on the body in the vertical direction is the acceleration due to gravity.