An arrow is projected by a bow vertically up with a velocity of 40 m/s, and reaches a target in 3 s. What is the velocity of the arrow just before it hits the target? How high is the target located?

Explanation:

Step one:

given data

initial velocity u= 40m/s

time taken t=3seconds

final velocity v=?

Step two:

applying the first equation of motion

v=u-gt---  (the -ve sign implies that the arrow is against gravity)

assume g=9.81m/s^2

v=40-9.81*3

v=40-29.43

v=10.57m/s

Step three:

how high the target is located

applying

s=ut-1/2gt^2

s=40*3-1/2(9.81)*3^2

s=120-88.29/2

s=120-44.145

s=75.86m

answer: good job it's all correct !

explanation: