Question

Q.
A
ball is projected with velocity 25m/s at an angle 37 degree with horizontal. find (A) total time of flight.
(B)time of ascent.
(C)time of descent.
(E)max. height.
(F)horizontal range.
(G)equation of trajectory.

Answer 1

Well, you have to assume the ground is level I guess... then
\[y _{f} = y _{i} + v _{yi}t + 1/2 a _{y}t {2}\]
will get you t since the initial and final heights are 0 and the initial velocity in the y direction is 25m/s sin 37 = 15m/s and a in the y direction is -9.8 m/s^2. To get the time of ascent, we know it will be half the time, but to calculate it, use
\[v _{yf} = v _{yi} + a _{y}t\]
where we know the final velocity is 0 because it has to stop before it will fall, the initial velocity is 15 m/s (see above) and the acceleration is still -9.8 m/s^2.
Time if decent is the same as the time of ascent, but you can calculate it using the same equation as above, remembering the initial velocity is now 0 and the final velocity is now -15 m/s (going downward).
You can get max height from \[v _{yf}^{2} = v _{yi}^{2} + 2 a _{y} \Delta y\] remembering initial height is 0, final velocity is 0 and using the acceleration of gravity, -9.8 m/s^2
The range can be had using the range equation, or by using the time from a) and \[\Delta x = v _{x}t\] where the x-velocity is 25 m/s cos 37 = 20 m/s
The path is a parabola -