Problem 1: A projectile is launch from a high plateau so that it just clears a c
ID: 1659238 • Letter: P
Question
Problem 1: A projectile is launch from a high plateau so that it just clears a cliff's edge and falls to the valley floor below. A person standing on the valley floor sees that is takes 4.5 s from the moment it passes the cliff's edge to when it strikes the valley floor, and that it strikes the ground 150m from the edge of the cliff with an impact speed of 105 m/s. Projectile Path Cliff's Edge Valley Floor Plateau 150 m a) How tall is the cliff (h)? b) At what initial velocity (magnitude and direction) was the projectile launched? c) How far from the cliff's edge was the projectile launched (R)? d) How long was the projectile in the air from its initial launch?Explanation / Answer
given, time taken for the ball to reach form the cliff to the ground, t = 4.5s
horizontal distance form the base of the cliff where the ball lands, d = 150 m
impact speed, |V| = 105 m/s
so let the initial speed of projectile be v and angle of projectile be theta
then from symmetry, when the ball passes the cliff, its speed is the same
so KE at this point + PE at this point = KE at the point when the ball touches the ground
0.5mv^2 + mgh = 0.5m|V|^2 .. (1)
also, it took time t to cover horizontal distance d
so, vcos(theta)*t = d
v*cos(theta)*4.5 = 150
v*cos(theta) = 33.333
and h = vsin(theta)*t + 0.5gt^2
h = v*sin(theta)*4.5 + 0.5*9.81*4.5^2
vsin(theta) = (h - 99.326)/4.5 = 0.2222h - 22.07244
squaring and adding
v^2 = 1598.281 + 0.04928h^2 - 9.8098h
so, (1598.281 + 0.04928h^2 - 9.8098h) + 2*9.81*h = 105^2
0.04928h^2 + 9.8101h - 9426.719 = 0
solving for h
h = 349.0144 m
a. height of cliff, h = 349.0144 m
b. intiial speed = v, ionitial angle = theta
v*cos(theta) = 33.333
vsin(theta) = 0.2222h - 22.07244 = 55.478
squaring and adding
v = 64.722 m/s
theta = 30.99 deg
c. Range of projectile = v^2*sin(2*theta)/g = 377.02 m
d. time of flight = R/vcos(theta) = 6.494 s
so the projectile was in air for 6.494 + 4.5 = 10.994 s