Answer to Question #160973 in Mechanics | Relativity for jireh

Question

aa projectile is launched at initial velocity of 12m/s and angle of 27 degree. (a) what is the maximum height reached by the projectile? (b) what is the time taken by the projectile? show your solution.

Answer 1

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

$v_y=v_{0y}-gt,$ $0=v_0sin\theta-gt$ $t=\dfrac{v_0sin\theta}{g}=\dfrac{12\ \dfrac{m}{s}\cdot sin27^{\circ}}{9.8\ \dfrac{m}{s^2}}=0.55\ s.$

Then, we can find the maximum height reached by the projectile:

$y_{max}=v_0tsin\theta-0.5gt^2,$ $y_{max}=12\ \dfrac{m}{s}\cdot0.55\ s\cdot sin27^{\circ}-0.5\cdot9.8\ \dfrac{m}{s^2}\cdot(0.55\ s)^2=1.51\ m.$

b) We can find the total flight time of the projectile from the formula:

$t_{flight}=2t=2\cdot0.55\ s=1.1\ s.$