Question

A boy launched from a waterslide into the air at an angle of 49º. With no wind, he would travel 430m horizontally where he lands in a pool. The time he is in the air is 2.9s.

-Find the initial horizontal velocity at launch?  

-What is the magnitude of the velocity at launch?

-Find his vertical velocity at landing?

-What is the magnitude of the velocity at landing?

-At maximum height he experiences a brief gust of wind that instantaneously reduces his horizontal velocity by 3.9m/s. What distance does he travel horizontally?

Answer 1

Range = 430 m

time of flight = 2.9 s

Using the equation

X = Ux*t + 0.5at^2

but acceleration in the horizontal direction is zero

X = Ux*t + 0

430 = Ux*2.9

Ux = 148.3 m/s

initial horizontal velocity = 148.3 m/s

since the projectile is launched at 49^o

Ux = U cos 49

148.3 = U*cos 49

U = 226 m/s

magnitude of velocity at launch = 226 m/s

vertical velocity at landing = 226 sin 49 (downwards direction)

= -170.57 m/s

magnitude of velocity at landing will be the same as the initial velocity = 226 m/s