/sci/ - Science & Math

[Anonymous No.4325882]

Hai /Sci/. Im doing some Kinematics work today to further expand my understanding of the topic and am stuck on this current question:

2. A ball was thrown from a balcony above a horizontal lawn. The velocity of projection was 10 ms-1 at an angle of elevation α, where tan α = ¾. The ball moved freely under gravity and took 3 s to reach the lawn from the instant when it was thrown. Calculate:

a) the vertical height above the lawn from which the ball was thrown

b) the horizontal distance between the point of projection and the point A at which the ball hit the lawn

c) the angle, to the nearest degree, between the direction of the velocity of the ball and the horizontal at the instant when the ball reached A.

I've easily worked out A and B but im having difficulties understanding C. Could anyone provide some assistance?

For the record (a) = 26.1 m and (b) = 24m.

Any help which points me in the right direction is appreaciated =]

[Anonymouse !70ikdWM2E6 No.4325883]

Reported.

[Anonymous No.4325886]

>>4325883

2.Homework threads will be deleted, and the poster banned.

i missed the part where this was homework. did i mention it was homework in any of the posts?

[Anonymous No.4325899]

I think, you need to know the vertical and horizontal velocity ... and the rest is easy

[Anonymous No.4325915]

>>4325899

you need to find the initial angle the ball was thrown at. so, if you know the initial velocity, and you figure out the magnitude of the initial velocity's x & y components, you can figure out the angle that the ball was thrown at.

[Anonymous No.4325916]

>>4325899

Thanks! I worked out vertical speed as 23.4 and horizontal as 8, used tan and its right.

thanks =]