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\(^{-2}\))
5J
10J
15J
20J
Explanation
Video Explanation
Post your Contribution
Discussions (16)
Reason it like this
At max hight above thr ground, P.E is max = mgh
at ground level, K.E is max = ½mv²
at any time between the maximum height and ground level, total energy possessed by the ball is a sum of K.E ans P.E ( ½mv² + mgh)
therefore
m= 0.1kg
v=10m/s
h=10m
g=10m/s²
T.E = mgh + ½mv²
T.E = (0.1×10×10) + (½×0.1×10²)
T.E = 10 + 5 = 15J
nothing sweet when you dey pack am clear for myschool app.
myschool admins more power to your elbow 💪💪💪
To find the total mechanical energy of the ball, we consider both:
Kinetic Energy (KE) due to motion
Potential Energy (PE) due to height
Given:
Mass,
Initial speed,
Height of tower,
Step 1: Kinetic Energy at the start
KE = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.1 \times 10^2 = 0.05 \times 100 = 5 \, \text{J}
Step 2: Potential Energy at the start
PE = mgh = 0.1 \times 10 \times 10 = 10 \, \text{J}
Step 3: Total Mechanical Energy
\text{Total Energy} = KE + PE = 5 + 10 = \boxed{15 \, \text{J}}
So, the total energy of the ball is 15 joules.
