A ball of mass 0.1kg is thrown vertically upwards with a speed of 10ms-1 from the top of a tower 10m high. Neglecting air resistance, its total energy just before hitting the ground is
(take g = 10ms-2)
(take g = 10ms-2)
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option C
Explanation:
Height h = 1/2 (U\(^2\)/g) = 10\(^2\)/20 = 5m
Energy before hitting ground is = P.E at the height of 5m.
P.E = mgh = 0.1 x 10 x 10= 10Â J
 K.E = 1/2 MV\(^2\) = 5J
T.E = K.E + P.E = 5Â + 10Â = 15J
Height h = 1/2 (U\(^2\)/g) = 10\(^2\)/20 = 5m
Energy before hitting ground is = P.E at the height of 5m.
P.E = mgh = 0.1 x 10 x 10= 10Â J
 K.E = 1/2 MV\(^2\) = 5J
T.E = K.E + P.E = 5Â + 10Â = 15J