Question

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?

m/s

(b) At what speed was the second ball thrown?

m/s

Answer #1

(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).

therefore -

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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