physic
posted by Chan on .
Hi! Question:A ball is tossed from an upperstory window of a building.The ball is given an initial velocity of 7.80m/s at an angle of 21.0 degrees below the horizontal.It strikes the ground 3.00s later.
A.How far horizontally from the base of the building does the ball strik the ground?
B.Find the height from which the ball was thrown.
C.How long does it take the ball to reach a point 10.0m below the level of launching? Thank you so much for your help

A. Multiply the horizontal component of the initial velocity by the time that the ball is in the air.
7.80 x cos 21 x 3.00 = ?
B. Solve the following equation for vertical height above the ground, for the unknown height H.
y = 0 = H 7.80 sin 21 t  (g/2) t^2
using t = 3.0 seconds. g is the acceleration of gravity, 9.8 m/s^2.
C. Use a different vertical height equation, with t as the unknown and the vertical height above the ground being
y = H 10
= H 7.80 sin 21 t  (g/2) t^2
Note that the H cancels out, leaving
10 = 7.80 sin 21 t  (g/2) t^2
There will be two roots. Take the one that is positive