A toga cap is thrown upward with a velocity of 3 m/s. Compute for its maximum height reached by the toga cap.
Answers
Answer:
(a) Irrespective of the direction of the motion of the ball, acceleration (which is actually acceleration due to gravity) always acts in the downward direction towards the centre of the Earth.
(b) At maximum height, velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value i.e., 9.8m/s
2
.
(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
(d) Initial velocity of the ball, u = 29.4 m/s
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero)
Acceleration, a =g=9.8m/s
2
From third equation of motion, height (s) can be calculated as:
v
2
−u
2
=2gs
s=(v
2
−u
2
)/2g= ((0)
2
−(29.4)
2
)/2(−9.8)=3s
Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the players hands = 3+3=6 s.