Question 71: Motion in 2D & Vectors

Question

A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

Accepted Answer

To determine how high a person can throw a ball given the maximum range, we need to use the principles of projectile motion in physics. The maximum range of a projectile is given by the formula:
$$ R = \frac{{v_0^2 \sin(2	heta)}}{g} $$
where:
$$ R $$ is the range
$$ v_0 $$ is the initial velocity
$$ 	heta $$ is the angle of projection
$$ g $$ is the acceleration due to gravity (approximately $$ 9.8 \, 	ext{m/s}^2 $$)
The maximum range is achieved when $$ 	heta = 45^\circ $$, thus $$ \sin(2	heta) = \sin(90^\circ) = 1 $$.
Given that the maximum range $$ R = 100 \, 	ext{m} $$, we can rewrite the range formula as:
$$ 100 = \frac{{v_0^2 \cdot 1}}{9.8} $$
Solving for $$ v_0^2 $$:
$$ v_0^2 = 100 	imes 9.8 = 980 $$
Next, the maximum height $$ H $$ reached by the ball can be calculated using the vertical component of the velocity. The formula for the maximum height is:
$$ H = \frac{{v_0^2 \sin^2(	heta)}}{2g} $$
Here, for a vertical throw, $$ 	heta = 90^\circ $$, and thus $$ \sin(90^\circ) = 1 $$.
Substituting the values, we get:
$$ H = \frac{{v_0^2 \cdot 1}}{2 \cdot 9.8} $$
Using $$ v_0^2 = 980 $$, we have:
$$ H = \frac{980}{2 	imes 9.8} = \frac{980}{19.6} = 50 \, 	ext{m} $$
Therefore, the correct answer is:
Option B: 50 m

Answer

25 m

Answer

50 m

Answer

100 m

Answer

200 m

Motion in 2D & Vectors

A person can throw a ball upto a maximum range of 100 m. How high above the ground he can throw the same ball?

Choose the correct answer: