Our website migration is now complete. If you experience any errors in pages, or if certain features are not functioning appropriately for you, we can fix it. Kindly contact us by DM, Email or Call.
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)
5 J
10 J
15 J
20 J
Explanation
Video Explanation
No video available
Post your Contribution
Discussions (153)
M= 0.1kg, V=10m, g=10m
Total Energy = K.E + P.E
i.e, 1/2MV^2 = mgh
i.e, (1/2 x 0.1 x 10^2) + (0.1x10x10)
i.e, 0.05 x 10^2= [5+10] = 15 J
K.E=1/2MV2 therefore, 1/2*0.1*100=5J
P.E=Mgh Therefore 0.1*10*10=10J
Total energy before hitting the ground=P.E+K.E I.e 10J+5J=15J.
M= 0.1, V= 10m/s, g=10m/s^2
Total Engergy = K.E + P.E
i.e::: 1/2mv2 + mgh = Total Energy.
by inputin d values, we have,
1/2×0.1x10^2 + 0.1x10x10
= 5 + 10, i.e 15J.
If u =10ms-2 using v2=u2-2ah, h=5m total height is 5+10=15m and E=mgh= 0.1*10*15=15j
The height of the tower is not taken into consideration. The answer should be 15J
Total energy just b4 hitting d ground=K.E+P.E wia k.e=(mv^2)/2 + mgh . . .subtituting ,Total energy=(0.1*10*10)/2 + 0.1*10*10 ==>10/2 + 10 = 5+10 = 15J
Given
M=0.1kg ,V=10m/s,g=10
Total energy before the bullet hit the ground =[(1/2)*M*(Vsq)] [M*G*H]
substitutin d values=[(1/2)*0.1*10sq] [0.1*10*10]=>
[5] [10]=15J
What is the method for solving heat problem involving two or more substances e.g calorimeter and water. Help plz
so I'm Currently reading this in the year 2026 and if this site still exists in the next decade and any New student is planning to write utme just know........ I don't know