Question

Incorrect Question 5 0/1 pts If the ball leaves the projectile launcher at a speed of 4.00 m/s at an angle of 300, and the projectile launcher is on a table at a height of 1.2 m, how far horizontally will it hit the ground? 2.56 m 5.12 m 1.98 m 3.33 m Use time from previous problem. Us Vx and time to get Xt.

Answer 1

First of all , we have to see the height above the table the ball is able to reach.

So, applying third equation of motion in vertical direction

v^{​​​​​​2}​-u² =2as

Here v=0 ( at the highest point the velocity is zero)

u= 4sin30

u= 2m/s

a=-9.81

so,

4=2*9.81*h

h=0.20387 meters

So the height above the ground will be,

H= 1.2+0.20387

H=1.40387 meters

Now, the time in which it reaches the maximum height be t^{​​​​​​}_{​​​​​​1}

So, applying first equation of motion,

v=u+at

2=9.81*t_{​​​​​​1}

t_{​​​​​​1} =0.20387 seconds

Now the time taken from the highest point to ground be t_{​​​​​​2}

applying second equation of motion,

H=ut +0.5at²

1.40387= 0.5*9.81*t₂​​​​​​²

t_{​​​​​​2}​​​=0.535 seconds.

So,

Total time taken will be

T=0.20387+0.535

T=0.73887 seconds

So in this time horizontal distance travelled will be

x= ut

Here, u is the initial speed in horizontal direction so,

x=(4cos30)*0.73887

x=2.559 or 2.56 m