Question
At time t = 0, a projectile is launched from ground level. At t = 2.00 s, it is displaced d = 53 m horizontally and h = 54 m vertically above the launch point. What are the (a) horizontal and (b)
vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?
vertical components of the initial velocity of the projectile? (c) At the instant it reaches its maximum height above ground level, what is its horizontal displacement D from the launch point?
Answers
u = 53/2 = 26.5 m/s forever
v = Vi - 9.81 t
h = Vi t - 4.9 t^2 = 54 =Vi(2)-4.9(4)
or
54 = 2 Vi - 19.6
so
Vi = 36.8 m/s
==========================
at top, v = 0
0 = 36.8 - 9.81 t
t = 3.75 seconds at top
but we know u = 26.5 m/s
so
D = u t = 26.5 * 3.75
v = Vi - 9.81 t
h = Vi t - 4.9 t^2 = 54 =Vi(2)-4.9(4)
or
54 = 2 Vi - 19.6
so
Vi = 36.8 m/s
==========================
at top, v = 0
0 = 36.8 - 9.81 t
t = 3.75 seconds at top
but we know u = 26.5 m/s
so
D = u t = 26.5 * 3.75