Answer to Question #176872 in Physics for Francine

Question

 An object has a launch velocity of 30.0 m/s at launch angle of 30.00relative to  

horizontal. Find the a) x and y component of initial velocity, b) time to reach the peak  

c) total time of flight, d) vertical height to reach the peak, and e) horizontal distance.

Answer 1

a) The horizontal (x) component of the velocity is:

$v_x = 30m/s\cdot \cos30\degree = 15\sqrt3 m/s\approx 13m/s$

The vertical (y) component of the velocity is:

$v_y = 30m/s\cdot \sin30\degree = 15m/s$

b) Time to reach the peak can be found as follows (https://en.wikipedia.org/wiki/Projectile_motion):

$t_{max} = \dfrac{v\sin\theta}{g} = \dfrac{30m/s\sin30\degree}{9.8m/s^2} \approx 1.53s$

where $g = 9.8m/s^2$ is the gravitational acceleration.

c)  The total time of flight is:

$t_{tot} = 2t_{max}\approx 3.06s$

d) Vertical height to reach the peak is:

$h_{max} = \dfrac{v^2}{2g} = \dfrac{(30m/s)^2}{2\cdot 9.8m/s^2}\approx 45.9m$

e) Horizontal distance is:

$d_{tot} = \dfrac{v^2\sin2\theta}{g} = \dfrac{(30m/s)^2\sin60\degree}{9.8m/s^2} \approx 79.5m$

Answer. a) 13m/s, 15m/s. b) 1.53s, c) 3.06s, d) 45.9m, e) 79.5m.