Can You Calculate the Required Brick Throwing Velocity to Hit a Moving Target?

[nice4life]

Homework Statement

You are on the top of a 100m high (approximately 30 story) building. You just finished building a brick grill so the tenants of the building can have barbecues. You have one brick left and do not want have to lug it back off the roof. Luckily, you see a garbage truck with its roof open (top of trash pile is 2m above the road) coming down a road that passes by the building. The road is 30m from the building at closest approach. The truck is moving down the road with constant velocity 15m/s. You need to make sure you hit the trashpile, or someone is going to get hurt. You can throw the brick comfortably at an angle of 45◦ above the horizontal.
(a) What velocity do you need to throw the brick so that it hits the top of the trash pile
when the truck is 30m away from the building? Does this sound like you could do it?

Homework Equations

deltax=Vox*t+1/2at^{^2} Vox = Initial velocity on x dimension.
Vy = Voysin0
Vx = Voxcos0

The Attempt at a Solution

I understand that I need Voysin0 and Voxcos0, but don't have Vo for x or y.
delta x = 30m and i think deltay = 98m because the truck is 2m about ground from the 100m building.

since on x axis/dimension accel. is constant 0. deltaX=Vox*t+0 but i don't know Vox.

Because they did not give me an initial velocity I am finding it very difficult to get this started. How do I find the Vsin0 and Vcos0 without relevant velocities?

 

[supratim1]

hello nice4life!
welcome to PF!

here, you are given the angle of throw = 45degree

and range is 30m.

vertical distance = 98m

so suppose initial velocity is V. so Vcos45.t = range, t = time of flight.

and Vsin45 is vertical component of velocity, use the equations of motions to calculate the time of flight, which is the time to get down 98m to the pile.

so solve two equations and you will get V.

 

[nice4life]

Thank you for the welcome, I am delighted I found this site.

so while i was working it all out i got t=5.11 and Vintial=8.3

i found this by two assumptions, 1) that Vcos45 and Vsin45 are the same 2) that it would take the same time for the brick to reach 30m distance and 98m high.

so i used
detlax=Vcos$$\Theta$$*t
30m=Vcos45*t
and
deltay=Vsin$$\Theta$$*t+1/2(a)t^2
98m=Vsin45*t+1/2(9.8)t^2
then set
Vcos45=30/t
plug it in
-98=(30/t)t-1/2(9.8)t^2
ALGEBRA...
t=5.11sec (i was happy when i got this )
So then
Vintital=(30/cos45*5.11)=8.3m/sdoes this seem right? Please may I get a tiny hint as to what I should do from here?

 

[supratim1]

you should use

98m=Vsin45*t - 1/2(9.8)t^2 instead of +

 

[nice4life]

i have worked it out a million times and i keep getting error. is delta y 98 or -98.

 

[supratim1]

its - 98