Question

Q.1) Projectile motion:A stone is thrown vertically into the air off the edge of a bridge 150ft above a river. If the initial velocity of the rock is 64ft/sec (upward0 and the accleration is constant -32ft/sec² (downward), find the following:

a. the velocity v(t) after t seconds.

b. The position s(t) of the stone above the river after t seconds.

c. The velocity at which the stone strikes the water.

Answer 1

the equations of projectil motion with constant aceleration in one dimension are:

1. 
2. 

in our case

•          (the initial position)
•        (the intial velocity)
•    (the aceleration)

A) Replacing the values in the equation (2):

or simply

B)Replacing the values in the equation (1):

or simply

C)If we want to calculate the velocity when the stone strikes the water () we need first calculate when en stone strikes the water (). To calculate that time we go to solve the equation (1) for :

the solve of that second grade equation is give by the formula

for our case( a = -16 , b = 64, c = 150)

the correct solution is obviously the positive . That is the time when the stone strikes the water, so the velocity of the strike is

the velocity is negative because goes downward