You throw a ball straight upward. As it leaves your hand, its speed is 15 m/s. (a) How much time does it take for the ball to reach the top of its trajectory? Start from a fundamental principle and show all your work.

the ball takes 1.53 s to reach the top of its trajectory.

Explanation:

given information:

the speed, v = - 15 m/s (moving upward)

(a) How much time does it take for the ball to reach the top of its trajectory?

we know that the speed for the vertical motion is

v = v₀ - gt, v₀ = 0

where

v = speed (m/s)

g = gravitational force (9.8 m/s²)

t = time (s)

thus

v = - gt

-15 = - 9.8 t

t = 15/9.8

 = 1.53 s

so, the time that is needed by the ball to reach the top its trajectory is 1.53 s

The ball takes 1.53 seconds to reach its top trajectory

Explanation:

The velocity of the ball will keep pushing it upwards until the velocity becomes zero. Therefore, the ball will reach the top of its trajectory when velocity i.e. V=0,

Fundamental principal of velocity is V = Vo + g*t

where, V=0  

Vo = 15 m/s

g = -9.8 m/s^2 (since ball is going upwards against the gravity)

t = ?  

0 = 15 + (-9.8 * t)

-15 = -9.8t

-15 / -9.8 = t

t = 1.53 Seconds

The ball takes 1.53 seconds to reach its top trajectory

e) none of the above answers are correct

unbalanced force

explanation: