JAMB - Physics (1991 - No. 3)
A small metal ball thrown vertically upwards from the top of a tower with an initial velocity of 20ms-1. If the ball took a total of 6s to reach ground level, determine the height of the tower [g = 10ms-2]
60m
80m
100m
120m
Explanation
Time from tower to top, t = \(\frac{u - v}{g}\)
= \(\frac{20 - 0}{10}\)
= 2s
distance from tower to top
x = ut - \(\frac{1}{2}\)gt2
= (20 x 2) - (\(\frac{1}{2}\) x 10 x 2 x 2)
= 20m
Time from the top to the ground, t = 6 - 2
= 4s
distance from top to ground,
x = ut - \(\frac{1}{2}\)gt2
= (0 x 4) + (\(\frac{1}{2}\) x 10 x 4 x 4)
= 80m
therefore, height of tower = 80 - 20
= 60m
= \(\frac{20 - 0}{10}\)
= 2s
distance from tower to top
x = ut - \(\frac{1}{2}\)gt2
= (20 x 2) - (\(\frac{1}{2}\) x 10 x 2 x 2)
= 20m
Time from the top to the ground, t = 6 - 2
= 4s
distance from top to ground,
x = ut - \(\frac{1}{2}\)gt2
= (0 x 4) + (\(\frac{1}{2}\) x 10 x 4 x 4)
= 80m
therefore, height of tower = 80 - 20
= 60m
Comments (0)
