Answer to Question #211156 in Physics for Michael

Question

An object is projected from the ground with an initial velocity of 25.0 m/s at 60 above the horizontal
Find the (a) horizontal and vertical components of the initial velocity, (b) time to reach the maximum,
(c) time of flight, (d) maximum height attained by the object, (e) range and (f) velocity upon striking the
ground.

Answer 1

(a) We can find the horizontal and vertical components of the initial velocity as follows:

$$v_{0x}=v_0cos\theta=25\ \dfrac{m}{s}\cdot cos60^{\circ}=12.5\ \dfrac{m}{s},$$ $$v_{0y}=v_0sin\theta=25\ \dfrac{m}{s}\cdot sin60^{\circ}=21.65\ \dfrac{m}{s}.$$

(b) We can find the time to reach the maximum height from the kinematic equation:

$$v_{y}=v_{0y}-gt,$$ $$0=v_{0y}-gt$$ $$t=\dfrac{v_{0y}}{g}=\dfrac{21.65\ \dfrac{m}{s}}{9.8\ \dfrac{m}{s^2}}=2.21\ s.$$

(c) We can find time of flight as follows:

$$t_{flight}=2t=2\cdot2.21\ s=4.42\ s.$$

(d) We can find the maximum height attained by the object from the kinematic equation:

$$y_{max}=v_{0y}t+\dfrac{1}{2}gt^2,$$ $$y_{max}=21.65\ \dfrac{m}{s}\cdot2.21\ s+\dfrac{1}{2}\cdot(-9.8\ \dfrac{m}{s^2})\cdot(2.21\ s)^2,$$ $$y_{max}=23.9\ m.$$

(e) We can find the range of the object from the kinematic equation:

$$x=v_{0x}t_{flight}=12.5\ \dfrac{m}{s}\cdot4.42\ s=55.25\ m.$$

(f) Let's first find the vertical component of the object's velocity just before it strikes the ground:

$$v_y=-gt=-9.8\ \dfrac{m}{s^2}\cdot2.21\ s=-21.6\ \dfrac{m}{s}.$$

Since the horizontal component of the object's velocity doesn't change during the flight, we can write:

$$v_x=v_{0x}=12.5\ \dfrac{m}{s}.$$

Finally, we can find the object's velocity upon striking the ground from the Pythagorean theorem:

$$v=\sqrt{v_x^2+v_y^2}=\sqrt{(12.5\ \dfrac{m}{s})^2+(-21.6\ \dfrac{m}{s})^2}=24.9\ \dfrac{m}{s}.$$