Answer to Question #147815 in Physics for Akeme edwina

Question

A stone is projected upwards at an angle of 30° to the horizontal from the top of a tower 100m and hits the ground at a point Q. If the initial velocity of projection is 100m/s2. Calculate the
I. Maximum height of the stone above the ground
II. Time it takes to reach this height
iII.time of flight
IV. Horizontal distance from the foot of the tower to the point Q. (Neglect air resistance and take g as 10m/s2

Answer 1

Let's write the equations of motion of the stone in horizontal and vertical directions:

$$x=v_0tcos\theta, (1)$$ $$y=y_0+v_0tsin\theta-\dfrac{1}{2}gt^2, (2)$$

here, $x$ is the horizontal displacement of the stone (or the horizontal distance from the foot of the tower to the point Q), $v_0=100\ \dfrac{m}{s}$ is the initial velocity of the stone, $t$ is the time of flight of the stone, $\theta=30^{\circ}$ is the launch angle, $y$ is the vertical displacement of the stone (or the height), $y_0=100\ m$ is the initial height from which the stone was projected and $g=10\ \dfrac{m}{s^2}$ is the acceleration due to gravity.

I) Let's first find the time that the stone takes to reach the maximum height from the kinematic equation:

$$v_y=v_0sin\theta-gt_{rise},$$ $$0=v_0sin\theta-gt_{rise},$$ $$t_{rise}=\dfrac{v_0sin\theta}{g}.$$

Then, we can substitute $t_{rise}$ into the second equation and find the maximum height:

$$y_{max}=y_0+v_0sin\theta\cdot\dfrac{v_0sin\theta}{g}-\dfrac{1}{2}g(\dfrac{v_0sin\theta}{g})^2,$$ $$y_{max}=y_0+\dfrac{v_0^2sin^2\theta}{2g},$$ $$y_{max}=100\ m+\dfrac{(100\ \dfrac{m}{s})^2\cdot sin^230^{\circ}}{2\cdot 10\ \dfrac{m}{s^2}}=225\ m.$$

II) We can find the time that the stone takes to reach the maximum height from the formula:

$$t_{rise}=\dfrac{v_0sin\theta}{g}=\dfrac{100\ \dfrac{m}{s}\cdot sin30^{\circ}}{10\ \dfrac{m}{s^2}}=5\ s.$$

III) Let's find the time of flight of the stone from the second equation:

$$0=100+100t\cdot sin30^{\circ}-\dfrac{1}{2}\cdot 10t^2,$$ $$5t^2-50t-100=0.$$

This quadratic equation has two roots:

$$t_1=\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{50+\sqrt{(-50)^2+2000}}{2\cdot 5}=11.7\ s,$$ $$t_2=\dfrac{-b-\sqrt{b^2-4ac}}{2a}=\dfrac{50-\sqrt{(-50)^2+2000}}{2\cdot 5}=-1.7\ s.$$

Since time can't be negative, the correct answer is $t=11.7\ s.$

IV) Finally, we can substitute the time of flight of the stone into the first equation and find the horizontal distance from the foot of the tower to the point Q:

$$x=v_0tcos\theta=100.0\ \dfrac{m}{s}\cdot 11.7\ s\cdot cos30^{\circ}=1013\ m.$$

Answer:

I) $y_{max}=225\ m.$

II) $t_{rise}=5\ s.$

III) $t=11.7\ s.$

IV) $x=1013\ m.$