Question

Determine the launch speed of a horizontally launched projectile that lands 28 meters from the base of a 21-meter high cliff

Answer 1

Answer: 13.5 m/s

Step-by-step explanation:

In order to solve this we must break this into horizontal and vertical dimensions.

We know these vertical quantities

Initial velocity vertical= 0 (since horizontally launched)

Acceleration vertical = -9.8 m/s^2 (acceleration due to gravity on earth)

Displacement vertical = - 21 m ( It will start 21 m on cliff and fall to the ground 0 m)

We know these horizontal quantities

Displacement horizontal= 28 m ( how far from the base we land)

Acceleration horizontal=0 ( we are in free fall)

We are looking for Horizontal velocity.

So we must use the kinematic equations to solve for this.

`displacement= Velocity_i*(Time)+ (1)/(2) *acceleration (time)^2`

we plug with the horizontal quantities and get

28= V( time) + 1/2*0*t^2 Since Acceleration is 0 we can cancel out the second part to get

28= V(time) Now we need to find time to solve for v

To do this we will use the same equation with the vertical quantities

`displacement= Velocity_i*(Time)+ (1)/(2) *acceleration (time)^2`

this will give us

-21=0 (time) +1/2 * -9.8* (time)^2 Now the first part will cancel because initial velocity is 0. This will give us;

-21= 1/2 * -9.8* (time)^2 now we solve for time and get

time = 2.07 s

Finally we can plug this in for the horizontal equation 28= V(time); this will give us

28= V( 2.07) We solve for v and get

Velocity = 13.5 m/s

Hope this helps