Question

A ball is thrown horizontally from the roof of a building 57 m tall and lands 40 m from the base. What was the ball's initial speed?

Answer 1

let v be the initial horizontal velocity..
suppose after time t the ball lands
then \[vt=40\] ............(1)
the initial vertical velocity of the ball is zero
we can use this equation...\[h _{y}=u _{y}t+1/2g t ^{2}\]
here \[u _{y}=0\] and \[h _{y}=57\]
then \[57=1/2g t ^{2}\]..................(2)
eliminate t from this two equation..you will get the anser..
\[v=11.7 m/s\]

Answer 2

THanks

Answer 3

welcome

Answer 4

if done close the question

Answer 5

An athlete executing a long jump leaves the ground at a 35° angle and travels 6.80 m.
) What was the takeoff speed?
(b) If this speed were increased by just 5.0%, how much longer would the jump be?

Answer 6

hint:draw a simple picture..let v be the velocity and t be the time..divide the velocity into two components...make two equation and find the unknown v by eliminating t..try it

Answer 7

when you say two components do you mean in the x direction and they direction

Answer 8

take a component horizontally (X component) another vertically(Y component)

Answer 9

but in order to find those, dont i need the velocity

Answer 10

let v be the velocity..the X component of the velocity is\[v \cos \theta\] and Y component is \[v \sin \theta \]
set two equation...and solve it

Answer 11

You know that in the x direction you traveled 6.80 m right?
So
\[v \cos \theta t = 6.80 m\]

We also know that the only acceleration in the y direction is gravity. The motion look like this.

The final velocity in the y direction is 0 right? So you need to substitute t with \[vsin \theta \] and solve for takeoff speed.