Question

A particle P is projected with speed 50m/s at an angle r to the horizontal from a point O on a horizontal plane. Given that the particle attains a maximum height of 31.25m, calculate:
(a) the value of r
(b) the speed of P after 3 seconds
(c) the horizontal distance travelled by P while above the height of 30m.

Answer 1

(a) Let's first find the $y$-component of the initial speed of the projectile:

$$v_y^2=v_{0y}^2-2gy_{max},$$ $$0=v_{0y}^2-2gy_{max},$$ $$v_{0y}=\sqrt{2gy_{max}}=\sqrt{2\cdot9.8\ \dfrac{m}{s^2}\cdot31.25\ m}=24.75\ \dfrac{m}{s}.$$

Then, we can find the angle $r$ from the trigonometry:

$$sinr=\dfrac{v_{0y}}{v_0},$$ $$r=sin^{-1}(\dfrac{v_{0y}}{v_0})=sin^{-1}(\dfrac{24.75\ \dfrac{m}{s}}{50.0\ \dfrac{m}{s}})=30^{\circ}.$$

(b) The horizontal component of particle's speed remains unchanged during the flight:

$$v_x=v_0cosr=50\ \dfrac{m}{s}\cdot cos30^{\circ}=43.3\ \dfrac{m}{s}.$$

Let's find the vertical component of particle's speed after 3 second:

$$v_y=v_0sinr-gt,$$ $$v_y=50\ \dfrac{m}{s}\cdot sin30^{\circ}-9.8\ \dfrac{m}{s^2}\cdot3\ s=-4.4\ \dfrac{m}{s}.$$

Finally, we can find the speed of P after 3 seconds from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2},$$ $$v=\sqrt{(43.3\ \dfrac{m}{s})^2+(-4.4\ \dfrac{m}{s})^2}=43.52\ \dfrac{m}{s}.$$

(c) Let's first find the time that the projectile takes to reach the maximum height:

$$0=v_0sinr-gt,$$ $$t_{max}=\dfrac{v_0sinr}{g}=\dfrac{50\ \dfrac{m}{s}\cdot sin30^{\circ}}{9.8\ \dfrac{m}{s^2}}=2.55\ s.$$

Then, let's calculate the time that the projectile takes to reach 30 meters height:

$$y=v_0tsinr-\dfrac{1}{2}gt^2,$$ $$4.9t^2-25t+30=0.$$

This quadratic equation has two roots $t_1=3.17\ s$ and $t_2=1.93\ s.$ Since, $t_1>t_{max}$ (which is impossible) the correct answer is $t=1.93\ s.$ Then, the total time during which the particle travelled above the height of 30 m can be calculated as follows:

$$t_{tot}=2(t_{max}-t)=2\cdot(2.55\ s-1.93\ s)=1.24\ s.$$

Finally, we can find the horizontal distance travelled by P while above the height of 30 m:

$$x=v_0tcosr=50\ \dfrac{m}{s}\cdot cos30^{\circ}\cdot1.24\ s=53.7\ m.$$