A ball is thrown straight upward and returns to the thrower's hand after 3.10 s in the air. A second ball thrown at an angle of 33.0° with the horizontal reaches the same maximum height as the first ball.
(a) At what speed was the first ball thrown?
(b) At what speed was the second ball thrown?
(a) Take the case of the first ball -
time to reach at the highest point, t = 3.10 / 2 = 1.55 s
(because 3.10 seconds is for it to go up and back down again)
initial velocity (u) = ?
acceleration due to gravity (a) = -9.8m/s^2 (negative because this is acting in the opposite direction that the ball is thrown)
final velocity when the ball is at the highest point (v) = 0m/s
use the expression -
v= u + at
u = v- at
u = 0 - (-9.8)*(1.55)
=> u = 15.2 m/s
(b) The vertical component of the velocity of the second ball must be the same as the initial velocity of the first ball (ie 15.2 m/s).
Lets assume that the speed of the second ball is (v)
v* sin(33) = 15.2
=> v = 15.2 / sin33 = 27.9 m/s.
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