Projectile Motions Question (1 Viewer)

[atobos281]

Seems extremely easy, but just cannot understand how to find 'v'.
Jack stands at the window of a building 6.2m above ground level. He throws his keys straight out of the window (horizontally) and hopes that his friend Tom, who is standing 10.4m out from the base of the building, will catch them. Ignoring air resistance and using g= 10 find the velocity at which Jack needs to throw his keys to 1dp. Q 14 from Maths in Focus.

Thankyou~

 

[jmk123]

Seems extremely easy, but just cannot understand how to find 'v'.
Jack stands at the window of a building 6.2m above ground level. He throws his keys straight out of the window (horizontally) and hopes that his friend Tom, who is standing 10.4m out from the base of the building, will catch them. Ignoring air resistance and using g= 10 find the velocity at which Jack needs to throw his keys to 1dp. Q 14 from Maths in Focus.

Thankyou~

Horizontal projection means theta=0. That should make things easier for u. And v=sqrt(Vx^2+Vy^2)

 

[atobos281]

Horizontal projection means theta=0. That should make things easier for u. And v=sqrt(Vx^2+Vy^2)

hmmm,... oh. Could u please tell me how to do the rest. Not too sure if it needs to be in Cartesian form or.. ty

 

[HeroicPandas]

hmmm,... oh. Could u please tell me how to do the rest. Not too sure if it needs to be in Cartesian form or.. ty

In every projectile motion question i do i derive the 6 equations of motions
Let v be the velocity of the particle

x'' = 0
x' = vcos (0) = v
x = vt

y'' = -10
y' = -10t + sin (0) = -10t
y = -5t^2 + 6.2

I urge u to please understand how to derive these important equations, if u cant find them, then u wont be able to do projectile motion questions