How do I do no 2, sorry for sideway image (1 Viewer)

[yashbb]

. S

 

[ExtremelyBoredUser]

no need to worry about horizontal component here since the ball is launched directly upwards. You can see this:

u = 12.5
ux = 12.5cos(90) = 12.5 * 0 = 0
uy = 12.5sin(90) = 12.5 * 1 = 12.5
angle is 90 degrees above horizontal.

u = 12.5 ms^-1, after converting kmh^-1 to ms^-1 (just divide by 3.6)

Assume positive to be upwards, negative to be downwards.

(a)
Peak height is when the velocity is 0

v^2 = u^2 + 2gs (since its in the vertical component form)

-u^2 = 2gs
therefore s = - u^2/2g = 7.97 m

(b)

v = u + gt = 12.5 - 9.8(0.5)
= 7.6 ms^-1

(c)

v = u + gt = 12.5 - 9.8(1.5)
= -2.2 ms^-1

 

[yashbb]

no need to worry about horizontal component here.

u = 12.5 ms^-1

(a)
Peak height is when the velocity is 0

v^2 = u^2 + 2gs (since its in the vertical component form)

-u^2 = 2gs
therefore s = - u^2/2g

(b)

v = u + gt = 12.5 - 9.8(0.5)

(c)

v = u + gt = 12.5 - 9.8(1.5)

once again, you sir, are just too smart.

 

[ExtremelyBoredUser]

once again, you sir, are just too smart.

If its confusing at first, just draw a diagram of whats going on and list all the known and unknown variables. Then you can see which kinematics equation will work best and solve the question.

 

[yashbb]

If its confusing at first, just draw a diagram of whats going on and list all the known and unknown variables. Then you can see which kinematics equation will work best and solve the question.

yep ok ty